Self Studies
Self Studies
Self Studies
Self Studies

Solutions Test 7

Result Self Studies

Solutions Test 7
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0

    Which of the following concentration factors is affected by change in temperature?

    Solution

    On changing temperature, solvent (H2O) is vaporised or condensed. Thus, volume of solution and solvent changes. Thus, any concentration factor dependent on volume is affected by change in temperature.
    (a) Molarity = Mole/Litre (Volume dependent)
    (b) Molality = Moles of solute/kg of solvent (weight dependent)
    (c) Mole fraction = ni/ntotal(unitless)
    (d) Weight fraction = wi/wtotal (unitless)

  • Question 2
    1 / -0

    6.02 x1020 molecules of urea are present in 100 mL of its solution. The concentration of urea solution is (N0 = 6.02 x 1023 mol-1)

    [AIEEE 2004]

    Solution

  • Question 3
    1 / -0

    Glucose solution is one molal. Glucose present in 1 kg glucose solution is

    Solution

    Glucose solution is one molal. Thus, 1000 g water (solvent) has glucose = one mole = 180 g

    Total mass of solution = 1000 +180 = 1180 g

    1180 g of solution has glucose = 180 g 

    ∴ 1000 g of solution has glucose =
    180 x 1000/1180 = 152.5 ≈ 153g  

  • Question 4
    1 / -0

    The molality of a urea solution in which 0.0100 g of urea (NH2CONH2) is added to 0.300 dm3 of water at STP is

    [AIEEE 2011]

    Solution

  • Question 5
    1 / -0

    A solution that is 20% ethanol by volume is found to have a density of 0.977 g/mL. Density of ethanol is 0.789 g/mL. Thus, mass per cent of ethanol solution is

    Solution

    20% ethanol by volume means 100 m l of (ethanol + H20 ) solution has ethanol = 20 mL 100 mL of solution = 100 x 0.977 g solution = 97.7 g solution 20 mL ethanol = 20 x 0.789 g ethanol = 15.78 g 

  • Question 6
    1 / -0

    A solution of NH4CI is prepared by dissolving 95 g of NH4CI in 200 g of H2O at 60° C. What is mass per cent when the solution is cooled to 20° C based on this

    Solution

    200 g H2O at 60°C has 95 g NH4Cl

    Based on solubility graph, at 20°C,

    100 g of H2O can dissolve = 36 g NH4CI

    hence, 200 g of H20 can dissolve = 72g  NH4CI

    Thus, NH4CI separated (crystallised) = 95 - 72 = 23 g

    Mass % = 36%

  • Question 7
    1 / -0

    Commercial concentrated nitric acid is 15.6 M. To prepare 10 L of 6.0 M nitric acid from it,

    Solution

    This is a case of dilution.

    Thus, 3.846 L of 15.6 M solution is diluted to 10.0 L by addition of 6.154 L of H20.

  • Question 8
    1 / -0

    AgNO3 sample is 85% by mass. To prepare 125 mL of 0.05 M AgNO3 solution, AgNO3 sample required is

    Solution

    125 mL of 0.05 M AgN03 = 125 x 0.05 millimoles 

  • Question 9
    1 / -0

    Mole fraction of C3H5(OH)3 in a solution of 36 gm of water and 46 gm of glycerine is 

    Solution

    The correct answer is option C
    The molar mass M2​ of non-volatile substance can be calculated from the following formula.

    Here, p0 vapour pressure of acetone and p is the vapour pressure of the solution.
    M1​ and W1​ are the molar mass and mass of acetone.
    M2​ and W2​ are the molar mass and mass of a non-volatile substance.
    Substitute values in the above expression.

    Hence, M2​=64.38g/mol.
    Thus, the molar mass of the non-volatile substance is 64 g/mol (approximately).
     

  • Question 10
    1 / -0

    Hardness of a water sample is 200 ppm CaCO3, Thus, molarity of CaCO3 is

    Solution

    Hard water has hardness = 200 ppm Thus, 106 mL(= 103 L)water has CaC03 =200 g =2 mol

  • Question 11
    1 / -0

    100 mL of aqueous solution of 0.01 M CaCI2 is evaporated to dryness when 0.15 g of residue is obtained. Thus, impurity present is

    Solution

    100mLof 0.01 M CaCI2 = 100 x 0.01 millimoles CaCI2
    = 0.001 mole CaCI2
    = 0.001 x 111 g CaCI2
    = 0.111 g pure CaCI2
    But, residue = 0.15 g
    Thus, impurity = 0.15 - 0.111= 0.039 g

     

  • Question 12
    1 / -0

    At 25 ° C, the density of 18 MH2SO4 is 1.8 g cm3. Thus, mass percentage of H2SO4 in aqueous solution is

    Solution

    18 MH2SO4 means18 mol L-1.
    1000 mL of H2S04 solution has H2SO4 = 18 mol
    = 18 x98g
    Also, 1000 m L = 1000 x 1.8 g H2SO4 solution
    Thus, (1000 x 1.8)g H2SO4 solution has H2SO4 =18 x 98 g

  • Question 13
    1 / -0

    Laboratory ammonia is 14.8 M NH3(ag)with a density of 0.8980 g/mL. Mole fraction of NH3 in this solution is

    Solution

    14.8 M NH3(ag ) means 14.8 moles NH3 in 1000 mL solution . 1000 mL solution = 1000 x 0,898 g solution

  • Question 14
    1 / -0

    An aqueous solution is 34% H3P04 by mass and has density 1.209 g mL -1 Molarity (I), molality (II) and normality (III) respectively, are

    Solution

    100g H3P04 aqueous solution has 34g H3P04

  • Question 15
    1 / -0

    If one assumes the volumes are additive, what is  in a solution obtained by mixing 275 mL of 0.200 M KN03, 325 mL of 0.40 M Mg(N03)2 and 400 mL H20?

    Solution

  • Question 16
    1 / -0

    How many grams of water would you add to 1.38 moles of CH3OH in 1 kg water to reduce the molality to 1.00 molal CH3OH(aq) ?

    Solution

    1.38 moles of CH3OH in w kg water

    1.38 moles of CH3OH in w Kg water 

    So that molarity = 1.0

  • Question 17
    1 / -0

    Three solutions have been provided.

    I. 3 g HCI in 1 kg H20.
    II. 2 L HCI(g) and 1 L H20 at room temperature.
    III. Aqueous solution of HCI with

    Solutions containing same number of moles of HCI is/are

    Solution

     

  • Question 18
    1 / -0

    Which of the following aqueous solutions has the highest concentration of Na+?

    Solution

  • Question 19
    1 / -0

    pH of Ba(OH)2 solution is 12. Number of millimoles present in 100 mL of Ba(OH)2 solution is

    Solution

  • Question 20
    1 / -0

    We have
    I. 25 mL of 1 M NaOH
    II. 10 mL of 0.50 M NaCI
    On mixing the two solutions, molar concentrations of Na+, OH- and Cl- respectively, are

    Solution

    This is a case of dilution.
    NaOH solution has been diluted by NaCI or vice-versa.
    Total volume = 35.0 mL

    M1V1 = M2V2

     

  • Question 21
    1 / -0

    A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed at 16 °C . Given,
    Density of H20 = 1 g mL-1
    Density of C2H,OH = 0.7939 gmL-1
    Density mixture = 0.9344 g mL-1
    Volume of the solution is

    Solution

    Volume can be additive or not depend on the nature of the solution

     

  • Question 22
    1 / -0

    How many grams of Mgl2 must be added to 250 mL of 0.0876 M Kl to produce a solution with [l-] = 0.10 M?

    Solution

     

  • Question 23
    1 / -0

    An aqueous solution is 6% methanol, CH3OH, by mass with d = 0.988 g mL-1. Thus, molarity of CH3OH in this solution is

    Solution

     (d) Aqueous 6% methanol solution by mass means 100 g of aqueous solution has = 6g CH3OH

  • Question 24
    1 / -0

    Matching List Type
    Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct

    An ethanol-water solution is prepared by dissolving 10.00 mL o f ethanol (C2H5OH, d = 0.789 g mL-1) in a sufficient volume o f w ater to produce 100 mL of solution with a density, d = 0.982 g mL-1. Match the concentration term in Column I with its value in Column II and select the answer from the code

    Solution

     

Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now