Question

Which equation shows the proper setup to find the volume of a 0.250 M CaCl2 solution that contains 39.5 g of CaCl2?

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Answer 1

Answer:

We know that molarity is the no of moles per unit volume.

From following relation we can calculate amount of cacl2 required (X)to prepare 0.25 M cacl2 250 ml solution.

1000ml—1M—111g cacl2

250ml —0.25—X g cacl2

So that from above

1000ml ×1M×X g= 250ml×0.25M×111g

X=[250ml×0.25M×111g]÷[1000ml×1M]

X=6.9375 g of cacl2

Hence by dissolving 6.9375 g of cacl2 in 250 ml volumetric flask with distilled water then you will get 0.25 M cacl2 solution.

Answer 2

Step by step explanation:

Given : The Molarity of $CaCl_2$  solution that contains $39.5 g$ of  $CaCl_2$.

To find : The volume of solution

Formula used : We use molarity formula,

$M=\frac{n}{V(s)(in L)}$

M is the molarity, n is the no. of moles of solute and V is the volume of solution in liters.

no. of moles of any compound 'a' are,

$n_a=\frac{W_a}{MM_a}$

W is the given mass and MM is molar mass of compound 'a'.

• Calculation for Volume of solution

We have molarity(M) of $CaCl_2$  solution and mass of $CaCl_2$ (W) are,

⇒  $Molarity=0.250M$   or   $Molarity = 0.250mol/L$

⇒  $W=39.5g$

As we know the molar mass of $CaCl_2$ ,

⇒  $MM=111g/mol$

by using formula the no. of moles are,

⇒  $n_a=\frac{W_a}{MM_a}$

⇒  $n=\frac{39.5g}{111g/mol}$

       $=0.356mol$

For finding volume we have to rearrange the formula of molarity,

⇒   $Molarity=\frac{n}{V(s)(in L)}$

taking  $V_s(in L)$ to L.H.S and molarity to denominator of R.H.S,

⇒  $V_s(in L)=\frac{n}{Molarity}$

We have n and Molarity as,

$n=0.356mol$    and  $Molarity = 0.250mol/L$

so the volume is,

⇒  $V_s(in L)=\frac{0.356mol}{0.250mol/L}$  

                 $=1.424L$

Hence we get the volume of solution as $1.424L$.