The equivalent weight of $H_{2} S O_{4}$ in the reaction is:
$N a O H + H_{2} S O_{4} \to N a H S O_{4} + H_{2} O$

49

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

98

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

40

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

80

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

Open in App

Solution

The correct option is B $98$
Equivalent weight $= \frac{M o l e c u l a r w e i g h t}{n f a c t o r}$

In the given reaction $H_{2} S O_{4}$ is acting as a monobasic acid. Therefore, n factor of $H_{2} S O_{4}$ is $1$

Equivalent Weight $\frac{98}{1} = 98$

Suggest Corrections

Similar questions

H_{2} S O_{4} + N a O H \to N a H S O_{4} + H_{2} O

H_{2} S O_{4}

H_{2} S O_{4} + N a O H \to N a H S O_{4} + H_{2} O

H_{2} S O_{4}

H_{2} S O_{4} + N a I \to N a_{2} S O_{4} + I_{2} + H_{2} S + H_{2} O

H_{2} S O_{4}

N a_{2} C r_{2} O_{7} + 3 S O_{2} + H_{2} S O_{4} \to 3 N a_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O

H_{2} S O_{4}

H_{2} S O_{4} + N a O H \to N a H S O_{4} + H_{2} O

Join BYJU'S Learning Program