Question

a 4.691 g sample of mgcl₂ is dissolved in enough water to give 750. ml of solution. what is the magnesium ion concentration in this solution? (3sf)
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integer, decimal, or e notation allowed
question 2
10 points
what mass of k₂co₃ is needed to prepare 200. ml of a solution having a potassium ion concentration of 0.150 m? (3sf)
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integer, decimal, or e notation allowed
question 3
10 points
the solubility of ba(no₃)₂ is 130.5 grams per liter at 0°c. how many moles of dissolved salt are present in 4.0 liters of a saturated solution of ba(no₃)₂ at 0°c? (2sf)

Explanation:

Step1: Calculate moles of \(MgCl_2\)

The molar mass of \(MgCl_2\) is \(M = 24.31+(2\times35.45)=95.21\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 4.691\space g\) and \(M = 95.21\space g/mol\), we have \(n=\frac{4.691}{95.21}\space mol\approx0.0493\space mol\).

Step2: Determine moles of \(Mg^{2 +}\) ions

Since \(MgCl_2
ightarrow Mg^{2+}+2Cl^{-}\), the mole ratio of \(MgCl_2\) to \(Mg^{2+}\) is \(1:1\). So, \(n(Mg^{2+})=n(MgCl_2) = 0.0493\space mol\).

Step3: Calculate the volume in liters

The volume \(V = 750\space mL=0.750\space L\).

Step4: Calculate the concentration of \(Mg^{2+}\) ions

Using the formula \(C=\frac{n}{V}\), where \(n = 0.0493\space mol\) and \(V = 0.750\space L\), we get \(C=\frac{0.0493}{0.750}\space M\approx0.0657\space M\).

Step1: Calculate moles of \(K^{+}\) ions

The volume \(V = 200\space mL = 0.200\space L\) and the concentration \(C(K^{+})=0.150\space M\). Using \(n = C\times V\), we have \(n(K^{+})=0.150\times0.200 = 0.0300\space mol\).

Step2: Determine moles of \(K_2CO_3\)

Since \(K_2CO_3
ightarrow2K^{+}+CO_3^{2 -}\), the mole ratio of \(K_2CO_3\) to \(K^{+}\) is \(1:2\). So, \(n(K_2CO_3)=\frac{n(K^{+})}{2}=\frac{0.0300}{2}=0.0150\space mol\).

Step3: Calculate the molar mass of \(K_2CO_3\)

The molar mass \(M(K_2CO_3)=(2\times39.10)+12.01+(3\times16.00)=138.21\space g/mol\).

Step4: Calculate the mass of \(K_2CO_3\)

Using \(m=n\times M\), where \(n = 0.0150\space mol\) and \(M = 138.21\space g/mol\), we get \(m=0.0150\times138.21\space g\approx2.07\space g\).

Step1: Calculate the mass of \(Ba(NO_3)_2\) in \(4.0\space L\) solution

The solubility is \(130.5\space g/L\). For \(V = 4.0\space L\), the mass \(m=130.5\times4.0 = 522\space g\).

Step2: Calculate the molar mass of \(Ba(NO_3)_2\)

The molar mass \(M(Ba(NO_3)_2)=137.33+(2\times14.01)+(6\times16.00)=261.35\space g/mol\).

Step3: Calculate the moles of \(Ba(NO_3)_2\)

Using \(n=\frac{m}{M}\), where \(m = 522\space g\) and \(M = 261.35\space g/mol\), we have \(n=\frac{522}{261.35}\space mol\approx2.0\space mol\).

Answer:

\(0.0657\)