A sample of seawater was found to contain 9.5 %w/w of MgCl2 and 5.85 %w/w of NaCl . Assuming complete ionization of salts, what will be the boiling point of sample. ( Kb for water =0.52 , at. wt. of Mg=24 , at. wt. of Na=23 , at. wt. of Cl=35.5 ) :-

Question

A sample of seawater was found to contain 9.5 %w/w of MgCl2 and 5.85 %w/w of NaCl . Assuming complete ionization of salts, what will be the boiling point of sample. ( Kb for water =0.52 , at. wt. of Mg=24 , at. wt. of Na=23 , at. wt. of Cl=35.5 ) :- (A) 101.5° C (B) 103.07° C (C) 100.52° C (D) 105.08° C

Tardigrade Admin · Accepted Answer

Consider 100gm of solution the mass of MgCl2=9.5gm & mass of NaCl=5.85gm Hence mass of solvent =100-(9 . 5 + 5 . 85)=84.65 for MgCl2,i1=3 & for NaCl,i2=2 (ÎT)b=(0 . 52 × (3 × (9 . 5/95) + 2 × (5 . 85/58 . 5))/84 . 65 × (10)- 3) (ÎT)b=(0 . 52 × (3 × 0 . 1 + 2 × 0 . 1)/84 . 65 × (10)- 3) =(0 . 52 × 0 . 5/84 . 65 × 10- 3)=3.07° C ∴ Boiling point of sea water sample =100+3.07 =103.07° C

Q. A sample of seawater was found to contain $9.5% w / w$ of $M g C l_{2}$ and $5.85% w / w$ of $N a Cl$ . Assuming complete ionization of salts, what will be the boiling point of sample. ( $K_{b}$ for water $= 0.52$ , at. wt. of $M g = 24$ , at. wt. of $N a = 23$ ,
at. wt. of $Cl = 35.5$ ) :-

$101. 5^{\circ} C$

$103.0 7^{\circ} C$

$100.5 2^{\circ} C$

$105.0 8^{\circ} C$

Q. A sample of seawater was found to contain 9.5%w/w of MgCl2​ and 5.85%w/w of NaCl . Assuming complete ionization of salts, what will be the boiling point of sample. ( Kb​ for water =0.52 , at. wt. of Mg=24 , at. wt. of Na=23 , at. wt. of Cl=35.5 ) :-

101.5∘C

103.07∘C

100.52∘C

105.08∘C

Q. A sample of seawater was found to contain $9.5% w / w$ of $M g C l_{2}$ and $5.85% w / w$ of $N a Cl$ . Assuming complete ionization of salts, what will be the boiling point of sample. ( $K_{b}$ for water $= 0.52$ , at. wt. of $M g = 24$ , at. wt. of $N a = 23$ ,
at. wt. of $Cl = 35.5$ ) :-

$101. 5^{\circ} C$

$103.0 7^{\circ} C$

$100.5 2^{\circ} C$

$105.0 8^{\circ} C$