The equivalent mass of $H_{3} B O_{3} (M = m o l a r m a s s o f H_{3} B O_{3})$ in its reaction with $N a O H$ to form $N a_{2} B_{4} O_{7}$ is equal to:

M

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

\frac{M}{2}

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

\frac{M}{4}

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

\frac{M}{6}

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

Open in App

Solution

The correct option is A $M$
$H_{2} B O_{3}$ is a mono basic acid.
So, $M =$ equivalent mass

Suggest Corrections

Similar questions

H_{3} B O_{3}

H_{3} B O_{3})

N a_{2} B_{4} O_{7}

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_{3} P O_{3}

H_{3} P O_{3} \to H_{3} P O_{4} + P H_{3}

I O_{4}^{-}

I_{2}

K I O_{3}

5 M H C l

Join BYJU'S Learning Program