6. you have to make 600 ml of a 0.40 m solution of sodium hydroxide. the only available solution is 2.0 m…

6. you have to make 600 ml of a 0.40 m solution of sodium hydroxide. the only available solution is 2.0 m. determine the amount needed to make the diluted solution.\n7. what is the molality of a solution in which 2.5 mol of ammonia, nh₃ is dissolved in 3 kg of water?\n8. what volume of 0.1 m hcl could be made with 3.7 moles of solute?\n9. how much solvent should be added to 9.5 grams of so₂ to prepare a 1.4 m solution?\n10. challenge question! 1.0 l of a solution is prepared by dissolving 125 g of kbr in it. if i took 150 ml of that solution and diluted it to 400 ml, determine the molarity of the diluted solution.

6. you have to make 600 ml of a 0.40 m solution of sodium hydroxide. the only available solution is 2.0 m. determine the amount needed to make the diluted solution.\n7. what is the molality of a solution in which 2.5 mol of ammonia, nh₃ is dissolved in 3 kg of water?\n8. what volume of 0.1 m hcl could be made with 3.7 moles of solute?\n9. how much solvent should be added to 9.5 grams of so₂ to prepare a 1.4 m solution?\n10. challenge question! 1.0 l of a solution is prepared by dissolving 125 g of kbr in it. if i took 150 ml of that solution and diluted it to 400 ml, determine the molarity of the diluted solution.

Answer

6.

Explanation:

Step1: Use dilution formula

$M_1V_1 = M_2V_2$, where $M_1$ is initial molarity, $V_1$ is initial volume, $M_2$ is final molarity and $V_2$ is final volume. Given $M_1 = 2.0\ M$, $M_2=0.40\ M$, $V_2 = 600\ mL$. We need to find $V_1$. $V_1=\frac{M_2V_2}{M_1}$

Step2: Substitute values

$V_1=\frac{0.40\ M\times600\ mL}{2.0\ M}=120\ mL$

Answer:

$120\ mL$

7.

Explanation:

Step1: Recall molality formula

Molality ($m$) is defined as $m=\frac{n}{m_{solvent}(kg)}$, where $n$ is the number of moles of solute and $m_{solvent}$ is the mass of the solvent in kilograms. Given $n = 2.5\ mol$ of $NH_3$ and $m_{solvent}=3\ kg$ of water.

Step2: Calculate molality

$m=\frac{2.5\ mol}{3\ kg}\approx0.83\ m$

Answer:

$0.83\ m$

8.

Explanation:

Step1: Recall molarity formula

Molarity ($M$) is defined as $M=\frac{n}{V}$, where $n$ is the number of moles of solute and $V$ is the volume of the solution in liters. We need to find $V$, and we know $M = 0.1\ M$ and $n=3.7\ mol$. $V=\frac{n}{M}$

Step2: Calculate volume

$V=\frac{3.7\ mol}{0.1\ M}=37\ L$

Answer:

$37\ L$

9.

Explanation:

Step1: Calculate moles of $SO_2$

The molar - mass of $SO_2$ is $M=(32 + 2\times16)\ g/mol=64\ g/mol$. The number of moles of $SO_2$, $n=\frac{m}{M}$, where $m = 9.5\ g$. So $n=\frac{9.5\ g}{64\ g/mol}\approx0.148\ mol$

Step2: Use molality formula to find mass of solvent

Molality ($m$) is $m = 1.4\ m$ and $m=\frac{n}{m_{solvent}(kg)}$. $m_{solvent}=\frac{n}{m}$ $m_{solvent}=\frac{0.148\ mol}{1.4\ m}\approx0.106\ kg = 106\ g$

Answer:

$106\ g$

10.

Explanation:

Step1: Calculate initial molarity of KBr solution

The molar - mass of $KBr$ is $M=(39 + 79.9)\ g/mol = 118.9\ g/mol$. The number of moles of $KBr$, $n=\frac{m}{M}=\frac{125\ g}{118.9\ g/mol}\approx1.05\ mol$ Initial molarity $M_1=\frac{n}{V}=\frac{1.05\ mol}{1.0\ L}=1.05\ M$

Step2: Use dilution formula

$M_1V_1 = M_2V_2$. Here $M_1 = 1.05\ M$, $V_1 = 150\ mL=0.15\ L$, $V_2 = 400\ mL = 0.4\ L$ $M_2=\frac{M_1V_1}{V_2}$ $M_2=\frac{1.05\ M\times0.15\ L}{0.4\ L}\approx0.394\ M$

Answer:

$0.394\ M$