Calculate volume of $0.2 M H_{2} {S O}_{4}$ required to react with following mixture is 12.5.
${N a O H (1 m o l) + C a (O H)}_{2} (2 m o l)$

True

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

False

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

Open in App

Solution

The correct option is A True
Moles of base or $O H^{⊖} = 1 m o l$ from $N a O H + 2 \times (2 m o l f r o m C a (O H)_{2})$
$2 m o l o f C a (O H)_{2}$
$= 5 m o l$
$∴ \frac{5}{2} m o l .$ of $H_{2} S O_{4}$ gives $5 m o l . H^{\oplus}$ which is what is required.
$5 / 2 = v o l u m e \times c o n c e n t r a t i o n$
$∴ V = \frac{50}{2 \times 0.2}$
$= 12.5 l$

Suggest Corrections

Similar questions

0.4

M N a O H

K H C_{2} O_{4} (1 m o l e) + H_{2} C_{2} O_{4} (2 m o l)

H C l

N a_{2} C O_{3}

N a H C O_{3}

0.4 M K M n O_{4}

K H C_{2} O_{4} (1 m o l) + H_{2} C_{2} O_{4} (2 m o l)

100 m l 0.1 M H_{2} O_{2}

400 m l 0.5 M H_{2} O_{2}

0.1 M K M n O_{4}

H_{2} O_{2}

Join BYJU'S Learning Program