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3 years ago

Question 15 How many grams of NaCl are required to make 500.0 mL of a 1.500 M solution? 58.40 g 175.3 g 14.60 g 43.83 g

1 answer:

4 0

Hi!

To make 500 mL of a 1,500 M solution of NaCl you'll require 43,83 g

To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:

$gNaCl=500mLsol* \frac{1L}{1000 mL}* \frac{1,500 mol NaCl}{1Lsol}* \frac{58,4428 g NaCl}{1 mol NaCl} \\ =43,83gNaCl$

Have a nice day!

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3 years ago

What are deltaTb and deltaTf for an aqueous solution that is 1.5g nacl in 0.250kg h2o? Given Kb=0.51 C/m and kr=1.86 C/m

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Answer:

$T_f$ for given question is 2.79 and $T_b$ is 0.52

$\Delta T_b = I \times K_b \times m$ {i- vant hoff’s constant ; Kb- constant ; m molarity }

M = no. of moles of the solute present in one kg of solution

Let the weight of amount of solute be “w” and its molecular mass be “M”

Let the mass of the solvent in the given question be “x”

$\Delta T_b = I \times K_b \times (w/M)/ x$

$\Delta T_b = I \times K_b \times w/Mx$

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3 years ago

In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +

Volgvan

Answer: The pH of the solution is 1.136

Explanation:

To calculate the moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For ammonia:

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol$

For nitric acid:

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

$0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol$

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

$NH_3+HNO_3\rightarrow NH_4NO_3$

At $t=0$ 0.0178 0.022

Completion 0 0.0042 0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

$pH=-\log[H^+]$

where,

$[H^+]=\frac{0.0042mol}{0.05741L}=0.0731M$

Putting values in above equation, we get:

$pH=-\log(0.0731)\\\\pH=1.136$

Hence, the pH of the solution is 1.136

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