In a 103 L sample of hard water- CaSO 4- MgSO 4 are present- If elevation in boiling point is 0-00208- C- then calculate the hardness of water in terms of ppm of CaCO 3- K b - for -H 2 O -0-52 Kkgmol -1

We know, Elevation in boiling point, Δ T_b=iK_b(m_1+m_2) ⇒0.00208=2 × 0.52 [ n_110^3+n_210^3 ] where n_1 and n_2 are the moles of CaSO_4 and MgSO_4 . ∴ n_1+n_2 ...

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Byju's Answer

Standard XII

Chemistry

Elevation in Boiling Point

In a 103 L sa...

Question

In a $10^{3} L$ sample of hard water, $C a S O_{4} & M g S O_{4}$ are present. If elevation in boiling point is ${0.00208}^{\circ} C$ , then calculate the hardness of water in terms of ppm of $C a C O_{3}$ . ( $K_{b}$ for $H_{2} O = 0.52 K k g m o l^{- 1}$ )

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Solution

We know,
Elevation in boiling point, $Δ T_{b} = i K_{b} (m_{1} + m_{2})$
$\Rightarrow 0.00208 = 2 \times 0.52 [\frac{n_{1}}{10^{3}} + \frac{n_{2}}{10^{3}}]$
where $n_{1}$ and $n_{2}$ are the moles of $C a S O_{4}$ and $M g S O_{4}$ .
$∴ n_{1} + n_{2} = \frac{2.08}{2 \times 0.52}$
$\Rightarrow n_{1} + n_{2} = 2$
$n$ factor of $C a S O_{4}$ and $M g S O_{4}$ are 2
Equivalent of $C a S O_{4} + M g S O_{4} = 2 \times 2 = 4$
Hardness of water is always expressed in ppm of $C a C O_{3}$
So, equivalent of $C a C O_{3} = 4$
$∴$ Mass. of $C a C O_{3} = (4 \times 50) g = 200 g$
Again, mass of water = $10^{3}$ L = $10^{3}$ kg = $10^{6}$ g
Hence, ppm $= \frac{200}{10^{6}} \times 10^{6} = 200$

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In a 103 L sample of hard water, CaSO4 & MgSO4 are present. If elevation in boiling point is 0.00208∘ C, then calculate the hardness of water in terms of ppm of CaCO3. (Kb for H2O=0.52 Kkgmol−1)

In a $10^{3} L$ sample of hard water, $C a S O_{4} & M g S O_{4}$ are present. If elevation in boiling point is ${0.00208}^{\circ} C$ , then calculate the hardness of water in terms of ppm of $C a C O_{3}$ . ( $K_{b}$ for $H_{2} O = 0.52 K k g m o l^{- 1}$ )