NEET Test 193

Lysosomes are membrane bound vesicles that contain digestive enzymes. They are single membraned and are isolated from their neighboring environment. The digestive enzymes present include proteases, lipases, nucleases, glycosidases and phosphatases. The lysosomes defend against bacterial and viral infection.

Therefore, all electrons are paired in O₂²⁻ and all others have an unpaired electron. Thus it is diamagnetic.

Nitrogen is one of the major constituents of proteins, nucleic acids, vitamins and hormones. It constituent cell membranes, certain proteins, all nucleic acids and nucleotide, and is required for all phosphorylation reactions.

Papaya plants has unisexual flowers.

Unisexual Flower: The flower has only one reproductive system. They can have a male reproductive system or a female reproductive system. Pollination is always cross.

Examples: Papaya, watermelon, corn.

The boiling point of a given compound is as follows:

I. Ethylmethylamine - 33 to 34 °C

II. Propylamine - 49 °C

III. Trimethylamine - 2.9 °C

The second group cations, copper Cu²⁺ and cadmium Cd²⁺ forms the cyano complexes: K₃[Cu(CN)₄] and K₂[Cd(CN)₄]. The second ionization of copper in the cyano complex is +1 and for cadmium it is +2 . Due to low ionization, the copper forms very stable complexes that do not dissociate easily in solution.

Here, the copper and cadmium can form complexes with the ligands. The tetra amine copper (II) sulfate and tetraamine cadmium (II) sulfate reacts with the potassium cyanide KCN. The KCN is the masking agent. The copper and cadmium form the cyano complexes. The complex is as follows:

[Cu(NH₃)₄]SO₄+KCN → K₃[Cu(CN)₄]

Similarly, the cadmium forms the cyano complex as follows:

[Cd(NH₃)₄]SO₄ +KCN → K₂[Cd(CN)₄]

When KCN is introduced in the solution, the cyano complexes are formed. When the solution is treated with the H₂ S gas, the unstable K₂[Cd(CN)₄] easily dissociates into the Cd²⁺ ions and precipitated as the CdS but K₃[Cu(CN)₄] is very stable complex therefore very few Cu²⁺ ions are present in the solution, therefore, it does not get precipitated as CuS Cd²⁺+ H₂ S ⇌ CdS(↓) + 2H +. 

Therefore, here K₃[Cu(CN)₄] is a more stable complex and K₂[Cd(CN)₄] is less stable complex.

Given:

Substitute 5 g for a mass of solute and 295 g for the mass of solvent and calculate the mass of solution.

As we know,

Mass of solution = mass of solute + mass of solvent

Mass of solution = 5 g +295 g = 300 g

Now, we have a mass solution and also we have given the density of the solution so, calculate the volume of the solution as follows:

Density = mass / volume 

volume = mass / Density 

Substitute 300 g for a mass of solution and 1.5g/cc for the density of the solution and calculate the volume of the solution.

volume of solution = 300 g / 1.5 g/cc

volume of solution = 200cc

Now, to calculate the moles of solute using the volume of solution and molarity of solution convert the volume of solution in L.

1 L= 1000cc

200cc × 1L/1000cc = 0.2 L

Substitute 0.05 M for molarity and 0.2 L for the volume of solution and calculate the moles of solute as follows:

Molarity = moles of solute / L of solution 

Moles of solute = Molarity × L of solution

Moles of solute = 0.05M × 0.2L = 0.01 mol

Now, we have moles of solute and mass of solute so calculate the molecular weight of solute as follows:

molecular weight = mass / mole 

Substitute 0.01 mol for moles of solute and 5 g for a mass of solution and calculate the molecular weight of the unknown solute.

Molecular weight = 5 g / 0.01 mol

Molecular weight = 5 g / 0.01 mol

Molecular weight = 500 g/mol

Thus, the molecular weight of the unknown solute is 500 g/mol.