Use the molar K2Cr2O7 mass (M) given by the supplier!

You can have the weight fraction of Cr (52) my dividing 52/M. The mass of the salt is larger than that of the element.

However, why are you preparing 50 mL of a stock solution? This volume is simply too small, the error on the weighted mass and on the volume are huge. And how will you dissolve the salt?

Good luck!

Dear Aarthy,

Based on my simple knowledge chemistry , Potassium dichromate VI (K2Cr2O7) is an oxidizing agent that oxidizes in aqueous solution to produce potassium cation (K+) and chromium anion (Cr2O72−). It has Molar mass of 294.185 g/mol

From your question, 50mg/ml (i.e. 0.05g in 1 ml)

To calculate the mass & % of Cr in chromate

= 2K + 2Cr + 7O2

39x2+2Cr+16x7 =294.185

Cr = 52g

2Cr= 104g

%2Cr in K2Cr2O7 = 35.4%

i.e for every 1g of K2Cr2O7 == 0.354g of Cr.

for 50mg (0.05g) Cr (vi) = 0.05/0.354 x 1

0.141g/ml

For 50ml == 0.141g x 50 = 7.062g.

I think your answer is correct. Regards

From the question the 50mg/ml is for Cr(VI) not the whole salt, which means wone has to dissolve more than that to get 50mg of Cr(VI) per given ml.

For me the required concentration is huge (50000 ppm which equals 50 mg/mL). First the required concentration should be 50 ug/mL (or 50 ppm). The dichromate is not the right choice as mentioned by some of the people above, as it is a strong oxidizing agent. chromium chloride or nitrate are much safer to be used for standard Cr solution preparation. The 50 ug/ml means 0.05 g of Cr dissolved in one liter. If you start with CrCl3 , calculate the amount of the salt which contains the 0.05 g by multiplying the 0.050 by the ratio CrCl3 (m.w. 158.5)/ (Cr a.w. 52). The result will be 0.1524 g provided the the salt is anhydrous. So 0.1524 g of anhydrous CrCl3 is dissolve in a little amount of water and then diluted to one liter with deionized water. Repeat the calculation if you have another salt of Cr in the same way. ... Good luck

Thank you for ur replies and suggestions. I need to prepare 50000 ppm (50 mg/ml) stock of Cr(VI) using K2Cr2O7. I used the following calc

104 g of Cr is in 294.19 g k2cr2o7

so, 1g (or1000 mg) cr will be in 2.82 g of k2cr2o7.

50 mg/ml Cr will be in 0.141 g/ml k2 cr2o7

Hence, for ppn of 50 ml of stock (containing 50 mg/ml cr),i need to dissolve 7.05 g of k2cr2o7 in 50 ml water

I need to know if this is right?@mahmood

Hi,

You may use the web tool - Mass Molarity Calculator to calculate the k2cr2o7 you need.

https://www.alfa-chemistry.com/mass-molarity-calculator.htm

For preparing stock solution(50 ml) containing 50 mg/ml of Cr(VI) you need to dissolve 7.0673 mg of K2Cr2O7 in 50 ml volumetric flask.

Remember the precautions needed for the manipulation of primary standards and dry your dichromate in oven at 110º.

For preparing stock solution(50 ml) containing 50 mg/ml of Cr(VI) you need to dissolve 7.0673 g of K2Cr2O7 in 50 ml volumetric flask.

valuable contributions

For preparing stock solution of 50 ppm of Cr(VI), you need to dissolve 7.0673 g of K2Cr2O7 in 50 ml D.W.

Regards.

How to prepare reagents for COD at 0～1500 mg/L high concentration water sample? I need a method of COD reagent preparation in detail (Using Hach DR3900 Spectrophotometer)

https://www.researchgate.net/post/How_to_prepare_50_ppm_mg_L_of_Chromium_from_K2Cr2O7/1

I think this would simply help.

Molar mass of K2Cr2O7 = 294.185 g/mol

Mass of Cr =52

mass of 2Cr = 2*52 = 104 g/mol

Therefore, to prepare 50 mg/ml in 50 ml...simply perform the operation below;

50mg/ml *(294.185 g/mol /104 g/mol)*50 ml

= 7071.755 mg.

Finally, weigh 7071.755 mg of K2Cr2O7 and dissolve in 50 ml of DI water.