Answer
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Hint: We can solve this problem with the help of formula of molarity because the concentration in the problem is asked moles per litre so Molarity =Number of moles /Volume of solution and we know that number of moles = mass of substance /Molar mass of substance .Here molar mass of nitric acid is $ = 1 + 14 + 3 \times 16 = 63$.
Complete step by step answer:
It is given in the problem statement that mass percent of nitric acid in the sample is $69\% $ and the density of the sample is $1.41gm{L^{ - 1}}$. So we can calculate from this the density of nitric acid which is $69\% $ of $1.41gm{L^{ - 1}}$. Hence density of nitric acid $ = 1.41 \times \dfrac{{69}}{{100}} = 0.9729gm{l^{ - 1}}$
We know that density $ = \dfrac{{Mass}}{{Volume}}$ , so density of nitric acid $ = $mass of nitric acid /volume $ \Rightarrow 0.9729 = \dfrac{{mass}}{{1000ml}}$ hence mass of $HN{O_3}$ $ = 972.9gm$
Now we will calculate the number of moles of nitric acid . Hence number of moles of nitric acid $ = $ mass of nitric acid /molar mass nitric acid $ = \dfrac{{972.9}}{{63}}$$ = 15.44$ moles
Now the molarity of the nitric acid =Number of moles of nitric acid /Volume of solution $ = \dfrac{{15.44}}{{1L}} = 15.44M$
Hence concentration of nitric acid in the sample is $15.44$moles per litre.
Note: We have approached this problem by finding the molarity of nitric acid in the sample which is expressed in moles per litre as the concentration of nitric acid in the question is asked in moles per litre. We have found out the number of moles of nitric acid by taking the ratio of its mass in grams to molar mass of nitric acid and then at last calculated molarity with the help of the formula of molarity.
Complete step by step answer:
It is given in the problem statement that mass percent of nitric acid in the sample is $69\% $ and the density of the sample is $1.41gm{L^{ - 1}}$. So we can calculate from this the density of nitric acid which is $69\% $ of $1.41gm{L^{ - 1}}$. Hence density of nitric acid $ = 1.41 \times \dfrac{{69}}{{100}} = 0.9729gm{l^{ - 1}}$
We know that density $ = \dfrac{{Mass}}{{Volume}}$ , so density of nitric acid $ = $mass of nitric acid /volume $ \Rightarrow 0.9729 = \dfrac{{mass}}{{1000ml}}$ hence mass of $HN{O_3}$ $ = 972.9gm$
Now we will calculate the number of moles of nitric acid . Hence number of moles of nitric acid $ = $ mass of nitric acid /molar mass nitric acid $ = \dfrac{{972.9}}{{63}}$$ = 15.44$ moles
Now the molarity of the nitric acid =Number of moles of nitric acid /Volume of solution $ = \dfrac{{15.44}}{{1L}} = 15.44M$
Hence concentration of nitric acid in the sample is $15.44$moles per litre.
Note: We have approached this problem by finding the molarity of nitric acid in the sample which is expressed in moles per litre as the concentration of nitric acid in the question is asked in moles per litre. We have found out the number of moles of nitric acid by taking the ratio of its mass in grams to molar mass of nitric acid and then at last calculated molarity with the help of the formula of molarity.
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Calculate the concentration of nitric acid in moles per litre in a sample which has a density $1.41gm{L^{ - 1}}$ and mass percent of nitric acid in it being $69\% $ .
Some Basic Concepts of Chemistry | NCERT EXERCISE 1.6 | Class 11 Chemistry Chapter 1 | Nandini Ma'am
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