Chemistry MCQs for NEET — Practice Questions with Answers

$\underset{0.1\mathrm{M}}{{\mathrm{AgNO}}_3}+\underset{0.2\mathrm{M}}{\mathrm{NaCl}}\stackrel{}{\to }\mathrm{AgCl}+{\mathrm{NaNO}}_3$

$0.1\mathrm{M}{\mathrm{AgNO}}_3$ reacts with 0.1 M NaCl to produce

0.1 M AgCl and $0.1\mathrm{M}{\mathrm{NaNO}}_3$

$\therefore {\mathrm{NO}}_3^{-}=\frac{0.1\mathrm{M}}{2}=0.05\mathrm{M}$

[∵ when equal volumes are mixed dilution occurs]  

Molarity $=\frac{\text{Wt.\hspace{0.17em}of\hspace{0.17em}solute}}{\text{Mol.\hspace{0.17em}wt.}}\times \frac{1000}{\text{Vol.\hspace{0.17em}of\hspace{0.17em}solution\hspace{0.17em}(in\hspace{0.17em}}ml.)}=\frac{98}{98}\times \frac{1000}{54.34}=18.4M$

$\left(\text{Vol.\hspace{0.17em}of\hspace{0.17em}solution\hspace{0.17em}}=\frac{{\displaystyle \text{mass}}}{{\displaystyle \text{density}}}=\frac{{\displaystyle 100}}{{\displaystyle 1.84}}=54.34ml\right)$   

Molecular mass of oxalic acid = 126

1000 ml of 1 M oxalic acid require oxalic acid = 126 g

$\therefore \text{\hspace{0.17em}\hspace{0.17em}}250ml$ of 1 M oxalic acid will require oxalic acid $=\frac{126}{1000}\times 250=31.5g$

Hence 250 ml of $\frac{M}{2}$ oxalic acid will require oxalic acid $=31.5\times \frac{1}{2}=15.75g$

∴ Mass of oxalic acid required = 15.75 g.  

In dilution, the following equation is applicable :

$M_1{V}_1$ = $M_2{V}_2$

$10MHCl$ = $5MHCl$

$10\times V_1=5\times 2.00$

$V_1=\frac{5\times 2.00}{10}L=1.00L$

Molarity of 95% $H_{2}S{O}_4=\frac{95}{98}\times \frac{1}{100/1.85}\times 1000$ = 17.93 M

Molarity of 15% $H_{2}S{O}_4=\frac{15}{98}\times \frac{1}{100/1.10}\times 1000$ = 1.68 M

$M_1{V}_1$ = $M_2{V}_2$

(95% H₂SO₄) = (15% H₂SO₄) 

$17.93\times V_1=1.68\times 100$ or V₁ = 9.4 cm³

$N=M\times \frac{\text{molecular mass (}{M}_{\text{2}}\text{)}}{\text{Equivalent mass (}E\text{)}}$

So, $M=N\times \frac{\text{Equivalent mass (}E\text{)}}{\text{Molecular mass (}{M}_{\text{2}})}$

Equivalent mass of salt $=\frac{\text{Molecular mass}}{\text{Total positive valency}}$

Equivalent mass of $N{a}_{2}C{O}_3=\frac{M_2}{2}$

$M=0.2\times \frac{M_2/2}{M_2}$; $M=\frac{0.2}{2}$ = 0.1 M 

H₂SO₄ is 95% by volume

Wt. of H₂SO₄ = 95 g

Vol. of solution = 100 ml

∴ moles of $H_{2}S{O}_4=\frac{95}{98}$ and weight of solution $=100\times 1.98=198g$

Weight of water $=198-95=103g$

Molality $=\frac{95\times 1000}{98\times 103}=9.412$

Hence molality of H₂SO₄ solution is 9.412  

Given H₂SO₄ is 93% by volume

Wt. of H₂SO₄ = 93 g

Volume of solution = 100 ml ∵ Density $=\frac{\text{mass}}{\text{volume}}$ 

$\therefore mass=d\times$volume

∴ weight of solution $=100\times 1.84g=184g$

wt. of water $=184-93=91g$

Molality $=\frac{\text{Moles}}{\text{wt. of water in }kg}=\frac{93\times 1000}{98\times 91}=10.42$

Mass of methanol = 16 g, Mol. mass of CH₃OH = 32

∴ No. of moles of methanol $=\frac{16}{32}=0.5moles$

No. of moles of water = $\frac{90}{18}=5moles$

∴ Mole fraction of methanol $=\frac{0.5}{5+0.5}=0.090$   

Since 18 g of water = 1mole

25 g of water = $\frac{25}{18}=1.38$ mole

Similarly, 46 g of ethanol = 1 mole

25 g of ethanol $=\frac{25}{46}=0.55$moles

Again, 60 g of acetic acid = 1 mole

50 g of acetic acid $=\frac{50}{60}=0.83$mole

∴ Mole fraction of water $=\frac{1.38}{1.38+0.55+0.83}=0.50$

Similarly, Mole fraction of ethanol $=\frac{0.55}{1.38+0.55+0.83}=0.19$  

Mole fraction of acetic acid $=\frac{0.83}{1.38+0.55+0.83}=0.3$