Question

1. prepare a 0.5m solution of nacl in 100 ml of di water (molar mass of nacl = 58.44g/mol) or 50 ml or 25 ml

Explanation:

Step1: Convert volume to liters

For 100 mL: \(V = 100\ mL\times\frac{1\ L}{1000\ mL}=0.1\ L\); for 50 mL: \(V = 50\ mL\times\frac{1\ L}{1000\ mL}=0.05\ L\); for 25 mL: \(V = 25\ mL\times\frac{1\ L}{1000\ mL}=0.025\ L\)

Step2: Calculate moles of NaCl

Using \(n = M\times V\), for 100 mL: \(n = 0.5\ mol/L\times0.1\ L = 0.05\ mol\); for 50 mL: \(n = 0.5\ mol/L\times0.05\ L = 0.025\ mol\); for 25 mL: \(n = 0.5\ mol/L\times0.025\ L = 0.0125\ mol\)

Step3: Calculate mass of NaCl

Using \(m=n\times M_{NaCl}\), for 100 mL: \(m = 0.05\ mol\times58.44\ g/mol = 2.922\ g\); for 50 mL: \(m = 0.025\ mol\times58.44\ g/mol = 1.461\ g\); for 25 mL: \(m = 0.0125\ mol\times58.44\ g/mol = 0.7305\ g\)

Answer:

We will calculate the mass of NaCl required for each volume separately.

For 100 mL solution:

1. First, convert the volume to liters:

• The volume \(V = 100\ mL=0.1\ L\).
• The molar - ity formula is \(M=\frac{n}{V}\), where \(M\) is molarity, \(n\) is the number of moles, and \(V\) is the volume of the solution in liters.
• We know \(M = 0.5\ mol/L\) and \(V = 0.1\ L\). Rearranging the formula for \(n\), we get \(n = M\times V\).
• \(n=0.5\ mol/L\times0.1\ L = 0.05\ mol\).
• The molar mass of \(NaCl\) is \(M_{NaCl}=58.44\ g/mol\).
• The mass \(m\) of \(NaCl\) is calculated using the formula \(m=n\times M_{NaCl}\).
• \(m = 0.05\ mol\times58.44\ g/mol=2.922\ g\).

For 50 mL solution:

1. Convert the volume to liters:

• \(V = 50\ mL = 0.05\ L\).
• Using \(n = M\times V\) with \(M = 0.5\ mol/L\) and \(V = 0.05\ L\), we have \(n=0.5\ mol/L\times0.05\ L = 0.025\ mol\).
• Then, \(m=n\times M_{NaCl}\), so \(m = 0.025\ mol\times58.44\ g/mol = 1.461\ g\).

For 25 mL solution:

1. Convert the volume to liters:

• \(V = 25\ mL=0.025\ L\).
• Using \(n = M\times V\) with \(M = 0.5\ mol/L\) and \(V = 0.025\ L\), we get \(n=0.5\ mol/L\times0.025\ L = 0.0125\ mol\).
• Then, \(m=n\times M_{NaCl}\), so \(m = 0.0125\ mol\times58.44\ g/mol=0.7305\ g\).

To prepare a \(0.5M\) solution of \(NaCl\):

• In 100 mL of DI water, dissolve \(2.922\ g\) of \(NaCl\).
• In 50 mL of DI water, dissolve \(1.461\ g\) of \(NaCl\).
• In 25 mL of DI water, dissolve \(0.7305\ g\) of \(NaCl\).