Chemistry Test - 4

Assume metal is M and mass of
metal is \(x\) g. As we know, Molecular weight of compound \(=2 \times\) vapour density \(=2 \times 50=100 g\)
\(\ldots \ldots\)(i)
Also, valency of metal is 2.
So, metal chloride is \(M C l_{2}\) Molecular weight of metal chloride \(=x+35.5 \times 2=x+71\)
\(=100 \quad\) (from equation i\()\)
So, molecular weight of metal, \(x\) \(=100-71=29 g / mol\)

As given, three moles of Mg require two moles of N. So, \(3 \times 24\) (molecular mass of Mg) mass of Mg will require \(2 \times 14\) (molecular mass of N) mass of N. So, N required for \(3.6 g\) of \(Mg\) \(=\frac{28}{72} \times 3.6=1.4 g\)

The number of moles of sulphur dioxide is one half the number of moles of carbon dioxide.
Hence, the number of moles of sulphur will be one half the number of moles of carbon.
Let the mixture contains \(x\) g of carbon and \(12-x\) g of sulphur. The number of moles of carbon and sulphur are \(\frac{x}{44}\) and \(\frac{12-x}{64}\)
respectively.
Hence, \(\frac{x}{44}=2\left(\frac{12-x}{64}\right)\)
\(\therefore x=5.14 g\)

The molar masses of \(N a_{2} S O_{4} .10 H_{2} O\) and \(H_{2} S O_{4}\) are \(322.2 g / mol\) and \(98 g / mol\)
respectively.
One molecule of \(N a_{2} S O_{4} .10 H_{2} O\) contains 14 oxygen atoms. One molecule of \(H_{2} S O_{4}\) contains 4 oxygen atoms.
Hence,
\(\frac{\text {mass of } N a_{2} S O_{4} .10 H_{2} O}{\text {molar mass of } N a_{2} S O_{4} .10 H_{2} O} \times 14 \times N_{A}=\frac{\text {mass of } H_{2} S O_{4}}{\text {molar mass of } H_{2} S O_{4}} \times 4 \times N_{A}\)
Substituting values in the above expression, we get \(\frac{1.61}{322.2} \times 14 \times N_{A}=\frac{x}{98} \times 4 \times N_{A}\)
\(x=1.78 gm\)

For every 100 g of organic compound, \(8 g\) of \(O\) and \(4 g\) of \(S\) are present.
The atomic weights of oxygen and sulphur are \(16 g / mol\) and \(32 g / mol\) respectively.
The molecule should contain at least one \(O\) atom and one \(S\) atom. The minimum possible molecular weight of compound is \(\frac{32}{4} \times 100=800 g / mol.\)

The equation is as follows:
\(3 H_{3} P O_{2} \rightarrow P H_{3}+2 H_{3} P O_{3}\)
3 moles of \(H_{3} P O_{2}\) gives 1 moles of \(P H_{3}\)
6 moles of \(H_{3} P O_{2}\) gives 2 moles of \(P H_{3}\)
Hence, 68 g of \(P H_{3}\) will be formed.

The relationship between the molarity and volume is,
\(M_{1} \times V_{1}=M_{2} \times V_{2}\)
Substituting values in the above expression, \(0.5 M \times 16 m l=0.20 M \times V_{2}\)
Hence, \(V_{2}=40 ml.\)
Hence, the volume of water required to make \(0.20 M\) solution from \(16 ml\) of \(0.5 M\) solution is \(40 m l-16 m l=24 m l\)

The correct stoichiometric coefficients can be obtained if we balance the chemical equation. The unbalanced chemical equation is \(A s_{4} O_{6}+S n C l_{2}+H C l \rightarrow A s_{4}+S n C l_{4}+H_{2} O\)
L.H.S contains 6 oxygen atoms. To balance oxygen atoms, the coefficient 6 is added to water on \(R.H.S.\)
Now, R.H.S contains \(12 H\) atoms. To balance \(H\) atoms, the coefficient \(r _{2}\) is added to \(HCl\) on the LHS.
Now, to balance chlorine, the coefficient 6 is added to \(S n C l_{2}\) on the \(L H S\) and the coefficient 6 is addd to \(Sn Cl _{4}\) on the \(RHS\). The balanced chemical equation is \(A s_{4} O_{6}+6 S n C l_{2}+12 H C l \rightarrow A s_{4}+6 S n C l_{4}+6 H_{2} O\)
Thus, the coefficients of the reactants are 1,6 and 12 respectively.

The weight of 2 nitrogen atoms is \(2 \times 14=28 g\)
\(100 g\) of aspartame contains 9.52 g of nitrogen.
Thus, the molecular weight of nitrogen is \(\frac{100}{9.52} \times 28=294.\)

Similar to the % labelling of oleum, \(P_{4} O_{10}+6 H_{2} O \rightarrow 4 H_{3} P O_{4}\)
\(127 \%\) mixture here means that \(27 g\) of \(H_{2}\) O reacts with \(100 g\) mixture to give \(127 g\) \(H_{3} P O_{4}\)
\(P_{4} O_{10}+6 H_{2} O \rightarrow 4 H_{3} P O_{4}\)
Moles of \(H_{2} O\) is \(\frac{27}{18}=1.5\) moles.
Now, moles of \(P_{4} O_{10}\) will be \(\frac{1.5}{6}=\frac{1}{4}\) moles of \(P_{4} O_{10}\)
\(\frac{1}{4} \times\) wt of \(P_{4} O_{10}=\) Mass of \(P_{4} O_{10}\)
\(\frac{1}{4} \times 284=71 g\) of \(P_{4} O_{10}\)