Molarity is defined as the number of moles of solute dissolved in 1 L of solution
the molarity of the solution to be prepared is 7.65 M
the volume to be prepared of CaBr₂ is 4.65 L
the number of CaBr₂ moles are - 7.65 mol/L x 4.65 L = 35.6 mol
the molar mass of CaBr₂ - 200 g/mol
mass of 35.6 mol - 35.6 mol x 200 g/mol = 7120 g
mass required of CaBr₂ to make the solution is 7120 g
Substance C is a acid.......
Answer:
Mass percent of food dyes = 0.0616%
Explanation:
Given data:
Mass of candy = 47.9 g
Calories = 240
Mass of fat = 10 g
Mass of carbohydrate = 34 g
Mass of protein = 2 g
Mass of food dyes = 29.5 mg
Mass percent of food dyes = ?
Solution:
First of all we will convert the mg into g.
Mass of food dyes = 29.5 mg × 1g /1000 mg = 0.0295 g
Mass percent of food dyes = mass of food dyes / total mass× 100
Now we will put the values.
Mass percent of food dyes = 0.0295 g / 47.9 g × 100
Mass percent of food dyes = 0.000616 × 100
Mass percent of food dyes = 0.0616%
<u>Given:</u>
Initial concentration of potassium iodate (KIO3) M1 = 0.31 M
Initial volume of KIO3 (stock solution) V1 = 10 ml
Final volume of KIO3 V2 = 100 ml
<u>To determine:</u>
The final concentration of KIO3 i.e. M2
<u>Explanation:</u>
Use the relation-
M1V1 = M2V2
M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M
Ans: The concentration of KIO3 after dilution is 0.031 M