Question

Arrange the following aqueous solutions in the order of their increasing boiling points
(i) 10^-4 M NaCl(ii) 10^-3 Urea
(iii) 10^-3 M MgCl2, (iv) 10^-3 M NaCl
(1) (i) <(ii) < (iv) < (iii)
(2) (ii) < (i) = (iii) < (iv)
(3) (i) < (ii) < (iii) < (iv)
(4) (iv) <(iii) < (i) = (ii)​

Kindly explain it plz.... and plz attach a photo.

Answer 1

Answer: (i) < (ii) < (iv) < (iii)

Explanation:

$\Delta T_b=i\times k_b\times m$

$\Delta T_b$ =  elevation in boiling point

i = Van'T Hoff factor  

$k_f$ = boiling point constant

m = molality

1. For $10^{-4}M$ $NaCl$

$NaCl\rightarrow Na^++Cl^-$  

i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be $10^{-4}+10^{-4}=2\times 10^{-4}$

2.  For $10^{-3}M$ urea

i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be $1\times 10^{-3}=10^{-3}$

3. For $10^{-3}M$ $MgCl_2$

$MgCl_2\rightarrow Mg^{2+}+2Cl^-$  

i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be $10^{-3}+2\times 10^{-3}=3\times 10^{-3}$

4. For $10^{-3}M$ $NaCl$

$NaCl\rightarrow Na^++Cl^-$  

i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be $10^{-3}+10^{-3}=2\times 10^{-3}$

Thus as concentration of ions follows the order: $3\times 10^{-3} MgCl_2$  >  $2\times 10^{-3} NaCl$  >$10^{-3}M urea$  >  $2\times 10^{-4} NaCl$  , the boiling point order will be: $2\times 10^{-4} NaCl$< $10^{-3}M urea$< $2\times 10^{-3} NaCl$   < $3\times 10^{-3} MgCl_2$

Thus order of their increasing boiling points is (i) < (ii) < (iv) < (iii)

Answer 2

Answer:hope u understand

Explanation: