Question

pls Ans this que now...............​

Answer 1

Explanation:

Mass percent of HNO3 in sample is 69 %

Thus, 100 g of HNO3 contains 69 g of HNO3 by mass.

Molar mass of HNO3

= { 1 + 14 + 3(16)} g.mol^{-1}g.mol

−1

= 1 + 14 + 48

= 63g\;mol^{-1}=63gmol

−1

Now, No. of moles in 69 g of HNO_{3}HNO

3

:

= \frac{69\:g}{63\:g\:mol^{-1}}

63gmol

−1

69g

= 1.095 mol

Volume of 100g HNO3 solution

= \frac{Mass\;of\;solution}{density\;of\;solution}

densityofsolution

Massofsolution

= \frac{100g}{1.41g\;mL^{-1}}

1.41gmL

−1

100g

= 70.92mL

= 70.92\times 10^{-3}\;L70.92×10

−3

L

Concentration of HNO3

= \frac{1.095\:mole}{70.92\times 10^{-3}L}

70.92×10

−3

L

1.095mole

= 15.44mol/L

Therefore, Concentration of HNO3 = 15.44 mol/L

Answer 2

Answer:

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