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

Question

Arrange the following aqueous solutions in the order of their increasing boiling points (i) 10-4 M NaCl (ii) 10-4 M Urea (iii) 10-3 M MgCl2 (iv) 10-2 M NaCl (A) (i) < (ii) < (iv) < (iii) (B) (ii) < (i) = (iii) < (iv) (C) (ii) < (i) < (iii) < (iv) (D) (iv) < (iii) < (i) = (ii)

Tardigrade Admin · Accepted Answer

For 10-4 M NaCl i = 2 10-4 M Urea i = 1 10-3 M MgCl2 i = 3 10-2 M NaCl i = 2 More the value of i, C, more will be the elevation in boiling point hence increasing order of boding point is 10-4 M Urea < 10-4 M NaCl < 10-3 M MgCl2 < 10-2 M NaCl

Q. Arrange the following aqueous solutions in the order of their increasing boiling points (i) 10−4MNaCl (ii) 10−4M Urea (iii) 10−3MMgCl2​ (iv) 10−2MNaCl

(i) < (ii) < (iv) < (iii)

(ii) < (i) = (iii) < (iv)

(ii) < (i) < (iii) < (iv)

(iv) < (iii) < (i) = (ii)

For 10−4MNaCli=2 10−4M Urea i=1 10−3MMgCl2​i=3 10−2MNaCli=2 More the value of i, C, more will be the elevation in boiling point hence increasing order of boding point is 10−4M Urea <10−4MNaCl<10−3MMgCl2​<10−2MNaCl

Q. Arrange the following aqueous solutions in the order of their increasing boiling points
(i) $1 0^{- 4} M N a Cl$
(ii) $1 0^{- 4} M$ Urea
(iii) $1 0^{- 3} M M g C l_{2}$
(iv) $1 0^{- 2} M N a Cl$