Suppose we have a 0.425 M solution. What volume (in mL) of this solution will be

Question

Suppose we have a 0.425 M solution. What volume (in mL) of this solution will be required to prepare 500 mL of a 0.184 M solution? 1150 mL 42.5 mL 39.1 mL 216 mL What mass of sodium nitrate is required to make 250. mL of a 0 476 M solution? 4.40 g 6.31 g 10.1 g 119 g Calcium reacts with hydrochloric acid to produce calcium chloride and hydrogen gas. What is the correct (balanced) form of the chemical equation that describes this 2Ca + 2HCl rightarrow 2CaCl + H_2 2Ca + HCl rightarrow Ca_2Cl + H Ca + 2HCl rightarrow CaCl_2 + H_2 Ca + HCl rightarrow CaCl + H

Explanation / Answer

(50) According to law of dilution   MV = M'V'

Where M = Molarity of stock = 0.425 M

V = Volume of the stock = ? mL

M' = Molarity of dilute solution = 0.184 M

V' = Volume of the dilute solution = 500 mL

Plug the values we get   , V = (M'V') / M

= 216 mL

(51) Number of moles , n = Molarity x volume in L

= 0.476 M x 250 mL x 10-3 L/mL

= 0.119 moles

Molar mass of NaNO3 is = 85 g/mol

So mass of NaNO3 required , m = number of moles x molar mass

= 0.119 mol x 85 g/mol

= 10.1 g

(52) Ca + 2HCl ----> CaCl2 + H2

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