Question

1. a solution contains 5.85g of sodium chloride dissolved in 5.00×10²ml of water. what is the molar concentration of the sodium chloride in mol/l? (ans 0.02mol/l)10. suppose that you are given a solution of 1.25mol/l sodium chloride in water, nacl. what volume must you dilute to prepare 50ml of 1.00mol/l nacl? (ans 40ml)11. find the ph of 1.5mol/l hydrochloric acid12. a balloon is filled with 2.50l of dry helium at 23.5°c. the balloon is placed in a freezer overnight. the next morning, the balloon is removed and the volume is found to be 2.15l. what was the final temperature of the balloon. (ans 255.42k)

Explanation:

Question 9

Step1: Calculate the molar mass of NaCl

The molar mass of \(Na\) is \(22.99\ g/mol\) and of \(Cl\) is \(35.45\ g/mol\). So, \(M_{NaCl}=22.99 + 35.45=58.44\ g/mol\)

Step2: Calculate the moles of NaCl

\(n=\frac{m}{M}\), where \(m = 5.85\ g\) and \(M = 58.44\ g/mol\). So, \(n=\frac{5.85}{58.44}\approx0.1\ mol\)

Step3: Convert the volume of water to liters

\(V = 5.00\times10^{2}\ mL=0.5\ L\)

Step4: Calculate the molar concentration

\(C=\frac{n}{V}\), where \(n = 0.1\ mol\) and \(V=0.5\ L\). So, \(C=\frac{0.1}{0.5}=0.2\ mol/L\)

Step1: Use the dilution formula \(C_1V_1 = C_2V_2\)

Here, \(C_1 = 1.25\ mol/L\), \(C_2=1.00\ mol/L\) and \(V_2 = 50\ mL\)

Step2: Solve for \(V_1\)

\(V_1=\frac{C_2V_2}{C_1}\)
Substitute the values: \(V_1=\frac{1.00\times50}{1.25}=40\ mL\)

Step1: For \(HCl\), \(HCl

ightarrow H^{+}+Cl^{-}\)
Since \(HCl\) is a strong acid, \(C_{H^{+}}=C_{HCl}\)
Given \(C_{HCl}=1.5\ mol/L\), so \(C_{H^{+}} = 1.5\ mol/L\)

Step2: Use the formula \(pH=-\log[H^{+}]\)

\(pH=-\log(1.5)\approx - 0.176\)

Answer:

\(0.2\ mol/L\)

Question 10