Question

what is the concentration of a solution which has 100.0g of lead(ii) nitrate dissolved in 500ml
mol/l
question 5 (1 point) retake question
if 250ml of 0.200 mol/l solution of magnesium chloride is mixed with 250 ml of 0.300 mol/l sodium hydroxide, magnesium hydroxide is a precipitate and the other product, sodium chloride, remains in solution. what is the concentration of sodium chloride solution?
mol/l

Explanation:

First Sub - Question (Concentration of lead(II) nitrate solution)

Step 1: Calculate moles of lead(II) nitrate

The molar mass of \(Pb(NO_3)_2\) is \(M = 207.2+(14.0\times2)+(16.0\times6)=331.2\space g/mol\). Moles \(n=\frac{m}{M}\), where \(m = 100.0\space g\). So \(n=\frac{100.0\space g}{331.2\space g/mol}\approx0.302\space mol\).

Step 2: Convert volume to liters

Volume \(V = 500\space ml=0.5\space L\).

Step 3: Calculate concentration

Concentration \(c=\frac{n}{V}\), so \(c=\frac{0.302\space mol}{0.5\space L}=0.604\space mol/L\) (approximate value, more accurately, \(\frac{100}{331.2\times0.5}=\frac{100}{165.6}\approx0.604\space mol/L\))

Second Sub - Question (Concentration of sodium chloride solution)

Step 1: Write the balanced chemical equation

The reaction is \(MgCl_2 + 2NaOH=Mg(OH)_2\downarrow+2NaCl\).

Step 2: Calculate moles of \(MgCl_2\) and \(NaOH\)

Moles of \(MgCl_2\): \(n_{MgCl_2}=c\times V = 0.200\space mol/L\times0.250\space L = 0.0500\space mol\)
Moles of \(NaOH\): \(n_{NaOH}=0.300\space mol/L\times0.250\space L = 0.0750\space mol\)
From the equation, 1 mole of \(MgCl_2\) reacts with 2 moles of \(NaOH\). The moles of \(NaOH\) required to react with \(0.0500\space mol\) of \(MgCl_2\) is \(2\times0.0500 = 0.100\space mol\), but we have only \(0.0750\space mol\) of \(NaOH\), so \(NaOH\) is the limiting reactant.

Step 3: Calculate moles of \(NaCl\) produced

From the equation, 2 moles of \(NaOH\) produce 2 moles of \(NaCl\). So moles of \(NaCl\) produced \(n_{NaCl}=n_{NaOH}=0.0750\space mol\) (since the ratio of \(NaOH\) to \(NaCl\) is 1:1 in the stoichiometry for production of \(NaCl\) from \(NaOH\) in this reaction).

Step 4: Calculate total volume of the solution

Total volume \(V_{total}=250\space ml + 250\space ml=500\space ml = 0.5\space L\)

Step 5: Calculate concentration of \(NaCl\)

Concentration \(c_{NaCl}=\frac{n_{NaCl}}{V_{total}}=\frac{0.0750\space mol}{0.5\space L}=0.150\space mol/L\)

Answer:

First sub - question: \(\approx0.604\space mol/L\) (or more accurately, \(\frac{100}{331.2\times0.5}\approx0.604\space mol/L\))
Second sub - question: \(0.150\space mol/L\)