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# Using the following information to help you answer parts (a) to (c):  $HN{O_3} + NaOH \to NaN{O_3} + {H_2}O$ RAM: $H = 1;Na = 23;O = 16;N = 14$ (A) If you have $0.55$ moles of $HN{O_3}$ how many oxygen atoms would you have?(B) If $0.365g$ of $NaOH$ is reacted with excess of $HN{O_3}$ what is the maximum mass of $NaN{O_3}$ which could be recovered?(C) $12.05c{m^3}$ of a $0.2065M$ aqueous $NaOH$ solution is titrated to the end-point by $25.05c{m^3}$ of $HN{O_3}$ solution. What is the concentration of the $HN{O_3}$ ?

Last updated date: 23rd Apr 2024
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Hint: Moles is the concept that represents the number of atoms present in one mole, which is equal to $6.023 \times {10^{23}}$ known as Avogadro’s number. As one mole consists of Avogadro's number of atoms, by multiplying the given number of moles with Avogadro’s number gives the number of atoms.

Given reaction is $HN{O_3} + NaOH \to NaN{O_3} + {H_2}O$
One mole consists of $6.023 \times {10^{23}}$ number of molecules, $0.55$ moles of $HN{O_3}$ consists of $6.023 \times {10^{23}} \times 0.55 = 3.312 \times {10^{23}}$ molecules. As there are three oxygen atoms in one molecule of $HN{O_3}$ .
The oxygen atoms will be $3 \times 3.312 \times {10^{23}} = 9.93 \times {10^{23}}$
Thus, $0.55$ moles of $HN{O_3}$ consist of $9.93 \times {10^{23}}$ oxygen atoms.
The number of moles is the ratio of mass to molar mass.
Given that the $0.365g$ of $NaOH$ is reacted, the molar mass of $NaOH$ is $40gmo{l^{ - 1}}$ .
The number of moles of $NaOH$ will be $\dfrac{{0.365}}{{40}} = 0.091moles$
As one mole of $NaOH$ reacts with one mole of $HN{O_3}$ to produce one mole of $NaN{O_3}$
Thus, $0.091moles$ of $NaOH$ produces $0.091moles$ of $NaN{O_3}$
The molar mass of $NaN{O_3}$ is $85gmo{l^{ - 1}}$
Thus, the mass of $NaN{O_3}$ will be $0.091mol \times 85gmo{l^{ - 1}} = 0.7335g$
If $0.365g$ of $NaOH$ is reacted with excess of $HN{O_3}$ the maximum mass of $NaN{O_3}$ which could be recovered is $0.7335g$
Given that the initial volume is $12.05c{m^3}$ with initial concentration of $0.2065M$ aqueous $NaOH$ solution is titrated to the end-point by a final volume of $25.05c{m^3}$ of $HN{O_3}$ solution.
Substitute these values in the formula, ${M_1}{V_1} = {M_2}{V_2}$
$12.05c{m^3} \times 0.2065M = {M_2} \times 25.05c{m^3}$
Further simplification, ${M_2} = 0.099M$
Thus, $12.05c{m^3}$ of a $0.2065M$ aqueous $NaOH$ solution is titrated to the end-point by $25.05c{m^3}$ of $HN{O_3}$ solution the concentration of the $HN{O_3}$ will be $0.099M$ .

Note:
While calculating the number of moles, the molar mass should be taken exactly and the balanced chemical equation only must be considered. Based on this balanced equation only, the mole ratio is taken and according to this the moles of product can be calculated.