(1)

Molarity = $\frac{w\times 1000}{M\times V\text{ (mL)}} ;$ $\frac{1}{20}=\frac{w\times 1000}{180\times 250}$

$w=\frac{180\times 250}{20\times 1000} ; w=2.25$$\mathrm{g}$

(2)

Molarity $=\frac{\text{Moles}}{\text{Volume (mL)}}\times 1000$

For HCl,

Number of moles $=\frac{M\times V\left(\text{mL}\right)}{1000}=\frac{0.75\times 25}{1000}=\frac{0.75}{40}$

Mass of HCl reacted $=\frac{0.75}{40}\times 36.5=0.684$$\mathrm{g}$

According to balanced equation

$\underset{1 \mathrm{mole} (40 \mathrm{g})}{\text{NaOH}}$          $+$             $\underset{1 \mathrm{mole} (36.5 \mathrm{g})}{\text{HCl}}$$\to \text{NaCl}+{\text{H}}_2\text{O}$  

36.5 g HCl reacts with 40 g NaOH

0.684 g HCl will react with $\frac{40}{36.5}\times 0.684=0.75$$\mathrm{g}$

NaOH left unreacted $=1-0.75=0.25\text{g or }250\text{mg}$

(4)

        $\underset{40+12+16\times 3=100g}{{\text{CaCO}}_3}$       $\stackrel{1200 K}{\to }$      $\underset{40+16=56\mathrm{g}}{\mathrm{CaO}}$     $+$      $\underset{44 \mathrm{g}}{{\text{CO}}_2}$

100 g ${\mathrm{CaCO}}_3$ produced ${\mathrm{CO}}_2$ gas = 44 g

20 g ${\mathrm{CaCO}}_3$ will produce ${\mathrm{CO}}_2$ gas = $\frac{44}{100}\times 20=8.8\text{g}$

If sample is 100% pure, ${\mathrm{CO}}_2$ produced = 8.8 g

If sample is 20% pure, ${\mathrm{CO}}_2$ produced = $\frac{8.8}{100}\times 20=1.76\text{g}$

(2)

Volume of HCl = 50 mL = 0.05 L

Molarity of HCl = 0.5 M

$\therefore$   Moles of HCl = $0.05\times 0.5=0.025$ moles

$CaC{O}_{\mathit{3}}+2HCl⟶ CaC{l}_{\mathit{2}}+C{O}_{\mathit{2}}+H_{\mathit{2}}O$

For 2 moles of HCl, $CaC{O}_{\mathit{3}}$ required = 1 mole

$\therefore$    For 0.025 moles of HCl, $CaC{O}_{\mathit{3}}$ required = $\frac{0.025}{2}$ moles

Mass of $CaC{O}_{\mathit{3}}$ required = $100\times \frac{0.025}{2}=1.25$ g

For 95% pure $CaC{O}_{\mathit{3}}$, mass of $CaC{O}_{\mathit{3}}$ required = $\frac{1.25}{95}\times 100\mathrm{g}=1.315\mathrm{g}\approx 1.32$ $\mathrm{g}$

(4)

Haber’s process, ${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$

2 moles of $N{H}_{\mathit{3}}$ are formed by 3 moles of $H_{\mathit{2}}$.

$\therefore$    20 moles of $N{H}_{\mathit{3}}$ will be formed by 30 moles of $H_{\mathit{2}}$.

(3)

Density = 1.28 g/cc, Conc. of solution = 2 M

Molar mass of NaOH = 40 g ${\mathrm{mol}}^{-1}$

Volume of solution = 1 L = 1000 mL

Mass of solution $=d\times V=1.28\times 1000=1280$ g

Mass of solute $=n\times \text{Molar mass}=2\times 40=80$ g

Mass of solvent = (1280 - 80) g = 1200 g

$\therefore$     Molality $=\frac{2\times 1000}{1200}=1.67$ m

(3)

[IMAGE 1]

${\mathrm{H}}_2\mathrm{O}$ gets absorbed by conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$. Gaseous mixture (containing CO and $C{O}_{\mathit{2}}$) when passed through KOH pellets, $C{O}_{\mathit{2}}$ gets absorbed.

Moles of CO left (unabsorbed) = $\frac{1}{20}+\frac{1}{20}=\frac{1}{10}$

Mass of CO = moles $\times$ molar mass = $\frac{1}{10}\times 28=2.8$$\text{g}$

(2)

16.9% solution of $AgN{O}_3$ means 16.9 g of $AgN{O}_3$ in 100 mL of solution $={\text{8.45 g of AgNO}}_3\text{ in 50 mL solution.}$

Similarly, 5.8 g of NaCl in 100 mL solution $\equiv \text{2.9 g of NaCl in 50 mL solution.}$

The reaction can be represented as:

[IMAGE 2]----------------------------------------------

$\therefore$    $\text{Mass of AgCl precipitated}=0.049\times 143.5=7.03\approx 7$$\text{g}$

(3)

$\underset{84 \mathrm{g}}{{\text{MgCO}}_{3\left(s\right)}}\stackrel{\Delta }{\to }\underset{40 g}{{\text{MgO}}_{\left(s\right)}}+{\text{CO}}_{2\left(g\right)}$

84 g of $MgC{O}_3\equiv$ 40 g of MgO

$\therefore$     ${\text{20 g of MgCO}}_3\equiv \frac{40}{84}\times 20=9.52\text{ g of MgO}$

Actual yield = 8 g of MgO

$\therefore$    $\%\text{purity}=\frac{8}{9.52}\times 100=84\%$

(1)

1 mole $\equiv$ 22.4 litres at STP.

$n_{{\text{H}}_2}=\frac{22.4}{22.4}=1\text{ mol}; n_{{\text{Cl}}_2}=\frac{11.2}{22.4}=0.5\text{ mol}$

Reaction is as,

[IMAGE 3]----------------------------

Here, $C{l}_2$ is the limiting reagent. So, 1 mole of ${\mathrm{HCl}}_{\left(\mathrm{g}\right)}$ is formed.