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Standard X

Chemistry

Volume of Gases and Number of Moles

200 mL soluti...

Question

200 mL solution of $N a O H$ and $N a_{2} C O_{3}$ was first titrated with N/10 $H C l$ in presence of $H P h$ 17.5 mL is required to end point. After this $M e O H$ was added and 2.5 mL of same $H C l$ is required. The amount of $N a O H$ in mixture is :

0.06 g per 100 mL

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0.06 g per 200 mL

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0.05 g per 100 mL

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0.012 g per 200 mL

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Solution

The correct option is A 0.06 g per 100 mL
200ml mixture of $N a O H$ & $N a_{2} C O_{3}$
Let $N a O H$ be x moles, $N a_{2} C O_{3}$ be y moles with hph indicator
we detect
x+y moles = $\frac{1}{10} \times \frac{17.5}{1000} = 1.75$ millimoles

next with methyl orange
y moles = $\frac{1}{10} \times \frac{2.5}{1000} = 0.25$ millimoles

⇒x=1.5 millimoles
y=0.25 millimoles

weight of $N a O H = 1.5 \times 40 \times 10^{- 3} = 60 m g$
Hence option A is correct

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