Question

If concentration of nitric acid is 70% (w/w) HNO3 then the mass of concentrated nitric acid required to prepare 200 mL of 1.0 M HNO3 is​

Answer 1

Answer:

The number of moles of nitric acid can be obtained by multiplying the molarity with volume

M×V=Moles of HNO  

3

​

=  

1000

250×2

​

=0.5

The mass of nitric acid can be obtained by multiplying the number of mols with molar mass and dividing with percentage concentration.

∴HNO  

3

​

 required =0.5×63×  

70

100

​

=45g

Thus 45.0 g conc. HNO  

3

​

 of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO  

3

​

.

Explanation:

Answer 2

Answer:

The mass of concentrated nitric acid required to prepare 200 mL of 1.0 M HNO₃ is​ 22.5g.

Explanation:

Given, the w/w concentration of nitric acid  = 70%

The molarity of HNO₃ = 1. 0 M

The volume of solution = 250ml = 0.25L

According to which one mole mole of  HNO₃  is dissolved in 1000ml of water.

We know the molecular mass of nitric acid = 63g/mol

We know, Molarity, $M=\frac{n}{V}$

where n is number of moles of nitric acid,

Number of moles of HNO₃ = Molarity × Volume of solution

Number of moles of HNO₃ = 1 × 0.25 = 0.25 moles

Mass of HNO₃ = 0.25 × 63 = 15.75g

The weight by weight concentration of HNO₃ = 70%

It means 70g of nitric acid in 100g of solution.

Therefore, mass of nitric acid required $=15.75\times \frac{100}{70}=22.5g$

                                                             

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