A sample of sea water was found to contain $9.5 %$ , w/w of $M g C l_{2}$ and $5.85 %$ , w/w of NaCl (Assuming complete ionization of salts) What will be the boiling point of sample

( $K_{b}$ for water $= 0.52$ ,K-kg/mol, At. Wt of Mg = 24, Na = 23, Cl = 35.5)

Question

A sample of sea water was found to contain $9.5 %$ , w/w of $M g C l_{2}$ and $5.85 %$ , w/w of NaCl (Assuming complete ionization of salts) What will be the boiling point of sample?

( $K_{b}$ for water $= 0.52$ ,K-kg/mol, At. Wt of Mg = 24, Na = 23, Cl = 35.5)

Smart Achievers · Accepted Answer

(2)

$In 100 gm solution$

$n_{{MgCl}_{2}} = 0.1$ $n_{NaCl} = 0.1$

$After dissociation total moles = 0.1 \times 3 + 0.1 \times 2 = 0.5$

$W_{water} = 100 - 9.5 - 5.85 = 84.65 gm$

$∴$ $Δ T_{b} = 0.52 \times 0.5 \times \frac{1000}{84.65} = 3.07 ° C$

$∴$ $T_{b} = 103.07 ° C$

Solution

Q.
A sample of sea water was found to contain $9.5 %$ , w/w of $M g C l_{2}$ and $5.85 %$ , w/w of NaCl (Assuming complete ionization of salts) What will be the boiling point of sample

( $K_{b}$ for water $= 0.52$ ,K-kg/mol, At. Wt of Mg = 24, Na = 23, Cl = 35.5)

1 $102.6 ° C$

2 $103.07 ° C$

3 $101.04 ° C$

4 $100.52 ° C$

Ans.
(2)

$In 100 gm solution$

$n_{{MgCl}_{2}} = 0.1$ $n_{NaCl} = 0.1$

$After dissociation total moles = 0.1 \times 3 + 0.1 \times 2 = 0.5$

$W_{water} = 100 - 9.5 - 5.85 = 84.65 gm$

$∴$ $Δ T_{b} = 0.52 \times 0.5 \times \frac{1000}{84.65} = 3.07 ° C$

$∴$ $T_{b} = 103.07 ° C$

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Solution

Q. A sample of sea water 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 ,K-kg/mol, At. Wt of Mg = 24, Na = 23, Cl = 35.5)

1 102.6°C

2 103.07°C

3 101.04°C

4 100.52°C

Ans. (2) In 100 gm solution nMgCl2=0.1 nNaCl=0.1 After dissociation total moles=0.1×3+0.1×2=0.5 Wwater=100-9.5-5.85=84.65 gm ∴ ΔTb=0.52×0.5×100084.65=3.07°C ∴ Tb=103.07°C