Solve for me from 4-7

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Here are the solutions to questions 4-7:

4. A solution of $\text{NaOH}$ contains 4.0g of $\text{NaOH}$ in 250ml of solution. Calculate the molarity of the solution. ($\text{Na}=23$, $\text{O}=16$, $\text{H}=1$)

Step 1: Calculate the molar mass of $\text{NaOH}$. $Molar mass of NaOH = (1 \times 23) + (1 \times 16) + (1 \times 1) = 23 + 16 + 1 = 40 g/mol$

Step 2: Calculate the number of moles of $\text{NaOH}$. $Moles of NaOH = \frac{Mass}{Molar mass} = \frac{4.0 g}{40 g/mol} = 0.10 mol$

Step 3: Convert the volume of the solution from milliliters to liters. $Volume of solution = 250 mL = 0.250 L$

Step 4: Calculate the molarity of the solution. $Molarity = \frac{Moles of solute}{Volume of solution (L)} = \frac{0.10 mol}{0.250 L} = \text{0.4 M}$

5. Calculate the mass of $\text{Na}_2CO_3$ required to prepare 250ml of 0.1M solution. ($\text{Na}=23$, $\text{C}=12$, $\text{O}=16$)

Step 1: Calculate the molar mass of $\text{Na}_2CO_3$. $Molar mass of Na_2CO_3 = (2 \times 23) + (1 \times 12) + (3 \times 16) = 46 + 12 + 48 = 106 g/mol$

Step 2: Convert the volume of the solution from milliliters to liters. $Volume of solution = 250 mL = 0.250 L$

Step 3: Calculate the number of moles of $\text{Na}_2CO_3$ required. $Moles of Na_2CO_3 = Molarity \times Volume (L) = 0.1 M \times 0.250 L = 0.025 mol$

Step 4: Calculate the mass of $\text{Na}_2CO_3$ required. $Mass of Na_2CO_3 = Moles \times Molar mass = 0.025 mol \times 106 g/mol = \text{2.65 g}$

6. Calculate the volume of 0.2M $\text{HCl}$ required to neutralize 25ml of 0.1M $\text{NaOH}$. ($\text{HCl} + NaOH \to NaCl + H_2\text{O}$)

Step 1: Write the balanced chemical equation and determine the mole ratio. The given equation is $\text{HCl} + NaOH \to NaCl + H_2\text{O}$. From the equation, 1 mole of $\text{HCl}$ reacts with 1 mole of $\text{NaOH}$. The mole ratio is $1:1$.

Step 2: Convert the volume of $\text{NaOH}$ solution from milliliters to liters. $Volume of NaOH = 25 mL = 0.025 L$

Step 3: Calculate the number of moles of $\text{NaOH}$. $Moles of NaOH = Molarity \times Volume (L) = 0.1 M \times 0.025 L = 0.0025 mol$

Step 4: Use the mole ratio to find the moles of $\text{HCl}$ required. Since the mole ratio of $\text{HCl}$ to $\text{NaOH}$ is $1:1$, $Moles of HCl = Moles of NaOH = 0.0025 mol$

Step 5: Calculate the volume of $\text{HCl}$ solution required. $Volume of HCl (L) = \frac{Moles of HCl}{Molarity of HCl} = \frac{0.0025 mol}{0.2 M} = 0.0125 L$ Convert to milliliters: $Volume of HCl = 0.0125 L \times 1000 mL/L = \text{12.5 mL}$

7. Calculate the percentage by mass of water of crystallization in $\text{Na}_2CO_3 \cdot 10H_2\text{O}$. ($\text{Na}=23$, $\text{C}=12$, $\text{O}=16$, $\text{H}=1$)

Step 1: Calculate the molar mass of $\text{Na}_2CO_3 \cdot 10H_2\text{O}$. $Molar mass of Na_2CO_3 = (2 \times 23) + (1 \times 12) + (3 \times 16) = 46 + 12 + 48 = 106 g/mol$ $Molar mass of H_2O = (2 \times 1) + (1 \times 16) = 2 + 16 = 18 g/mol$ $Molar mass of 10H_2O = 10 \times 18 = 180 g/mol$ $Total molar mass of Na_2CO_3 \cdot 10H_2O = 106 + 180 = 286 g/mol$

Step 2: Calculate the percentage by mass of water of crystallization. $Percentage by mass of H_2O = \frac{Mass of 10H_2O}{Total molar mass of Na_2CO_3 \cdot 10H_2O} \times 100\%$ $= \frac{180 g/mol}{286 g/mol} \times 100\% \approx \text{62.94\%}$

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