A solution contains 410.3 g of $H_{2} S O_{4}$ per litre of solution at $20^{o} C$ . If the density is $1.243 g / m L$ , what will be its molarity and molality?

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Solution

Mol. mass of $H_{2} S O_{4} = 98$

No. of moles of $H_{2} S O_{4} = \frac{410.3}{98} = 4.186$

Molarity of $H_{2} S O_{4}$ solution $= \frac{N o . o f m o l e s o f H_{2} S O_{4}}{V o l u m e o f s o l . i n l i t r e}$

$= \frac{4.186}{1} = 4.186 M$

Mass of 1 litre $H_{2} S O_{4}$ solution $= 1000 \times 1.243 = 1243 g$

Mass of water $= (1243 - 410.3) = 832.7 g = \frac{832.7}{1000} k g$

Molality of solution $= \frac{N o . o f m o l e s o f H_{2} S O_{4}}{M a s s o f w a t e r i n k g}$

$= \frac{4.186}{832.7} \times 1000$

$= 5.027 m$

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A solution contains 410.3 g of H2SO4 per litre of solution at 20oC. If the density is 1.243g/mL, what will be its molarity and molality?

A solution contains 410.3 g of $H_{2} S O_{4}$ per litre of solution at $20^{o} C$ . If the density is $1.243 g / m L$ , what will be its molarity and molality?