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Chapter 8: Problem 140

Upon heating a litre of semi-molar \(\mathrm{HCl}\) solution, \(2.675 \mathrm{~g}\) of hydrogen chloride is lost and the volume of the solution shrinks to \(750 \mathrm{ml}\). The molarity of resultant solution is (a) \(0.569 \mathrm{M}\) (b) \(0.5 \mathrm{M}\) (c) \(0.42 \mathrm{M}\) (d) \(1.707 \mathrm{M}\)

Short Answer

Expert verified
Answer: The molarity of the resulting solution after heating is 0.569 M.

Step by step solution

01

Calculate the moles of HCl in the initial solution

First, we need to find out how many moles of HCl were in the initial solution. We are given that the initial solution was semi-molar, which means its molarity was 0.5M. The moles of HCl in the initial solution can be found using the molarity equation mentioned above: \(M = \dfrac{n}{V}\) Rearranging to find n: \(n = M \times V\) Here, M = 0.5 M V = 1 L (given) \(n_{initial} = 0.5 M \times 1 L = 0.5\) moles So, the initial solution had 0.5 moles of HCl.
02

Calculate the mass loss in moles

We are given that 2.675 g of hydrogen chloride was lost when heated. To find the moles of HCl lost, we need to convert this mass into moles by dividing the mass by the molar mass of HCl. Molar mass of HCl = 36.5 g/mol. \(n_{lost} = \dfrac{Mass_{lost}}{Molar_{mass}}\) So, \(n_{lost} = \dfrac{2.675 g}{36.5 g/mol} = 0.07329\) moles
03

Calculate the moles of HCl left

Now, subtract the moles of HCl lost from the initial moles of HCl to find the moles of HCl left after heating: \(n_{final} = n_{initial} - n_{lost}\) So, \(n_{final} = 0.5 - 0.07329 = 0.42671\) moles
04

Calculate the final molarity of the solution

Finally, we can calculate the molarity of the final solution using the moles of HCl left and the final volume of the solution (750 ml = 0.75 L): \(M_{final} = \dfrac{n_{final}}{V_{final}}\) So, \(M_{final} = \dfrac{0.42671 \, moles}{0.75 L} = 0.569 M\) Therefore, the molarity of the resulting solution after heating is 0.569 M, which corresponds to answer choice (a).

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Most popular questions from this chapter

An organic compound contains \(40 \%\) carbon and \(6.67 \%\) hydrogen by mass. Which of the following represents the empirical formula of the compound? (a) \(\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{2} \mathrm{O}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (d) \(\mathrm{CH}_{3} \mathrm{O}\)

What mass of carbon disulphide, \(\mathrm{CS}_{2}\), can be completely oxidized to \(\mathrm{SO}_{2}\) and \(\mathrm{CO}_{2}\) by the oxygen liberated when \(325 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{O}_{2}\) react with water? (a) \(316.67 \mathrm{~g}\) (b) \(52.78 \mathrm{~g}\) (c) \(633.33 \mathrm{~g}\) (d) \(211.11 \mathrm{~g}\)

When \(20 \mathrm{~g} \mathrm{Fe}_{2} \mathrm{O}_{3}\) is reacted with \(50 \mathrm{~g}\) of \(\mathrm{HCl}, \mathrm{FeCl}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) are formed. The amount of unreacted \(\mathrm{HCl}\) is \((\mathrm{Fe}=56)\) (a) \(27.375 \mathrm{~g}\) (b) \(22.625 \mathrm{~g}\) (c) \(30 \mathrm{~g}\) (d) \(4.75 \mathrm{~g}\)

Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The mass percentage of carbon in cortisone is \(69.98 \%\). What is the molecular mass of cortisone? (a) \(180.05\) (b) \(360.1\) (c) \(312.8\) (d) \(205.8\)

Iron form two oxides. If for the same mass of iron, mass of oxygen combined in the first oxide is two-third of the mass of oxygen combined in the second oxide, the ratio of valency of iron in first and second oxide is (a) \(1: 1\) (b) \(2: 3\) (c) \(3: 2\) (d) \(2: 5\)
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