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Stoichiometry

10mL of 0.1...

Question

$10 m L$ of $0.1 M H_{2} {S O}_{4}$ is needed for complete reaction with $425 m L$ of $11.44 g^{- 1} {N a}_{2} {c C O}_{3} . x H_{2} O$ . The value of X.

110

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2033

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600

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54

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Solution

The correct option is A $110$
$m_{1} v_{1} = m_{2} v_{2}$
$10 m l \times 0.1 M = M_{2} \times 425 m l$
$\frac{10 \times 0.1}{425} = M_{2}; M_{2} = 0.002135$
Molar mass of $N a_{2} C O_{3} . x H_{2} O = 23 \times 2 + 14 + 15 \times 3 + 18 x = 46 + 12 + 48 + 18 x = 106 + 18 x$
Molarity $= \frac{\frac{11.44}{106 + 18 x}}{v}$
Therefore, $0.00235 = \frac{11.44}{106 + 18 x} \times \frac{425}{1000}$
$106 + 18 x = \frac{4.862}{0.00235} 106 + 18 x = 2068.93 x = 109.05$

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