Find the volume of water (in mL) added to 50 mL concentrated nitric acid $H N O_{3}$
$(d = 1.52 g m l^{- 1}$ , 69% by mass) that forms diluted nitric acid ( $d = 1.15 g m L^{- 1}$ , 19% by mass $H N O_{3}$ ).

Open in App

Solution

Let volume of dilute solution that can be formed is V.
Mass of $H N O_{3}$ will be same before and after dilution
So, $V \times 1.15 \times 0.19 = 50 \times 1.52 \times 0.69$
$⟹ V = 240$ mL
Volume of water added = 240 - 50 = 190 mL.

Suggest Corrections

Similar questions

Q. Find the volume of water (in mL) added to 50 mL concentrated nitric acid

H N O_{3}

(d = 1.52 g m l^{- 1}

, 69% by mass) that forms diluted nitric acid (

d = 1.15 g m L^{- 1}

, 19% by mass

H N O_{3}

Q. 240 ml of diluted nitric acid (d = 1.15 g/ml, 19% by mass)

H N O_{3}

is to be prepared from conc.

H N O_{3}

(d =1.52 g/ml, 69% by mass). How much water has to be added (in mL) to the concentrated solution to achieve this?

Q. Determine the volume of dilute nitric acid (

d = 1.11

m L^{- 1}

19 %

w/w

H N O_{3}

) that can be prepared by diluting with water

50

mL of conc.

H N O_{3}

(

d = 1.42

m L^{- 1}

69.8 %

w/w).

Q. Calculate the concentration of nitric acid in moles per litre in a sample which has a density.

1.41

m L^{- 1}

and the mass per cent of nitric acid in it being

69 %

Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.5 g per mL and the mass percent of nitric acid in it being 69%.

Join BYJU'S Learning Program

Find the volume of water (in mL) added to 50 mL concentrated nitric acid HNO3 (d=1.52 g ml–1, 69% by mass) that forms diluted nitric acid (d=1.15 g mL–1, 19% by mass HNO3).

Let volume of dilute solution that can be formed is V. Mass of HNO3 will be same before and after dilution So, V×1.15×0.19=50×1.52×0.69 ⟹V=240 mL Volume of water added = 240 - 50 = 190 mL.

Find the volume of water (in mL) added to 50 mL concentrated nitric acid $H N O_{3}$
$(d = 1.52 g m l^{- 1}$ , 69% by mass) that forms diluted nitric acid ( $d = 1.15 g m L^{- 1}$ , 19% by mass $H N O_{3}$ ).

Let volume of dilute solution that can be formed is V.
Mass of $H N O_{3}$ will be same before and after dilution
So, $V \times 1.15 \times 0.19 = 50 \times 1.52 \times 0.69$
$⟹ V = 240$ mL
Volume of water added = 240 - 50 = 190 mL.