Question

how much ca(no3)2 in mg must be present in 50 ml of a solution with 2.35ppm of ca​

Answer 1

Answer:

Explanation: For 2.35 ppm of Ca,

 10^6 ml = 2.35 g of Ca

We have to find for 50 ml

for 50  ml = 50/(10^6)*2.35

                  = 117.5*10^-6 g

Now, 164 g of Ca(NO3)2 = 40g of Ca

                                            = 117.5*10^-6 g of Ca.

                                            = 117.5*10^-6*164 g of Ca(NO3)2

                                             = 481.75 g

                                             = 481.75 *10^-3 mg

                                              = 0.48

Answer 2

Given:

Ca = 50 ml (2.35 ppm)

To Find:

Present Ca(No₃)₂ = ?

Solution:

For 2.35 ppm of Ca,

⇒  $2.35 g \ of \ Ca = 10^6 \ ml$

Now,

For 50 ml,

$=\frac{50}{10^6}\times 2.35$

$=117.5\times 10^{-6} \ g$

So that, 164 g of Ca(No₃)₂ will be,

$=40 g \ Ca$

$=117.5\times 10^{-6}g \ Ca$

$=117.5\times 10^{-6}\times 164g \ Ca(NO_{3})_{2}$

$=481.75\times 10^{-3}$

$=0.48 \ mg$

Therefore, the amount of Ca(No₃)₂ will be "0.48 mg".