The mass of $80 %$ pure $H_{2} S O_{4}$ required to completely neutralise $60 g$ of $N a O H$ is

183.75 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

58.8 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

73.5 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

98 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $183.75 g$
$80 % H_{2} S O_{4}$
$60 g m N a O H$
$H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O + 2 C O_{2}$
$\frac{x g m}{98 g m} m o l e s \frac{60}{40} = 1.5 m o l e s$
$\frac{x}{98} = 1.5$
$x = 98 \times 2.5$
$= g m$
Let $g m$ be impure sample containity
$196 g m$ $H_{2} S O_{4}$ which is $80 %$
$∵ y \times \frac{80}{100} = 196 = x$
$y = \frac{196 \times 100}{80}$
$= 245 g m$
$y = \frac{147 \times 100}{80}$
$183.75 g m$

Suggest Corrections

Similar questions

Mass of $70$ % $H_{2} S O_{4}$ solution (by mass) required for neutralisation of $1$ mol of $N a O H$ is:

70 %

H_{2} S O_{4}

1

N a O H

70 % H_{2} S O_{4}

N a O H

H_{2} S O_{4}

H_{3} P O_{4}

N a O H

Join BYJU'S Learning Program