You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Chemistry

Relationship between Vapor Density & Relative Molecular Mass

What volume o...

Question

What volume of HCl solution of density 1.2 g/ $c m^{3}$ and containing $36.5 %$ by weight HCl, must bw allowed to react with zinc (Zn) in order to liberate 4.0 g of hydrogen?

333.33 mL

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

500 mL

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

614.66 mL

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

None of these

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

Open in App

Solution

The correct option is A 333.33 mL
$Z n + 2 H C l \to Z n C l_{2} + H_{2}$ ;
moles of $H_{2}$ evolved $= 2$
$∴$ moles of $H C l$ required $= 4$
$∴ \frac{V \times 1.2 \times 0.365}{36.5} = 4$
$V = 333.33 m L$

Suggest Corrections

Similar questions

6.13

g sample of

K_{2} C r_{2} O_{7}

(

96 %

purity) is allowed to react with

320

mL of

H C l

solution having density

1.15

g/mL and containing

30 %

by mass of

H C l

. What mass of

C l_{2}

is liberated?

K_{2} C r_{2} O_{7} + 14 H C l ⟶ 2 K C l + 3 C r C l_{3} + 7 H_{2} O + 3 C l_{2}

Q. 36.5 (%w/w)

H C l

solution has density equal to

1.2 g m L^{- 1}

then :

29.2 % (w / w) H C l

stock solution has a density of

1.25 g m L^{- 1}

. The molecular weight of

H C l

36.5 g m o l^{- 1} .

The volume

(m L)

of stock solution required to prepare a

200 m L

solution of

0.4 M H C l

is:

Q. 29.2% (w/w)

H C l

stock solution has a density of

1.25 g m l^{- 1}

. The molecular weight of

H C l

36.5 g m o l^{- 1}

. The volume (ml) of stock solution required to prepare

200 m l

solution of

0.4

H C l

is:

Q. 48.67 % (w/W) HCl stock solution has a density of

1.5 g m L^{- 1} .

The molecular weight of HCl is

36.5 g m o l^{- 1} .

The volume (mL) of stock required to prepare 200 mL of 0.4 M HCl is: ___

Join BYJU'S Learning Program

What volume of HCl solution of density 1.2 g/cm3 and containing 36.5% by weight HCl, must bw allowed to react with zinc (Zn) in order to liberate 4.0 g of hydrogen?

The correct option is A 333.33 mLZn+2HCl→ZnCl2+H2;moles of H2 evolved =2 ∴ moles of HCl required =4 ∴V×1.2×0.36536.5=4V=333.33mL

What volume of HCl solution of density 1.2 g/ $c m^{3}$ and containing $36.5 %$ by weight HCl, must bw allowed to react with zinc (Zn) in order to liberate 4.0 g of hydrogen?

The correct option is A 333.33 mL
$Z n + 2 H C l \to Z n C l_{2} + H_{2}$ ;
moles of $H_{2}$ evolved $= 2$
$∴$ moles of $H C l$ required $= 4$
$∴ \frac{V \times 1.2 \times 0.365}{36.5} = 4$
$V = 333.33 m L$