How many grams of sodium chloride are present in a 0.75 m solution with a volume of 500.0 milliliters?

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219.375 g of NaCl is present in a 0.75 M solution with a volume of 500.0 milliliter

Explanation: -

Volume of NaCl solution = 500 mL

= 500/1000 = 0.5 L

Strength of NaCl solution = 0.75 M

Number of moles of NaCl = Strength of NaCl x Volume of NaCl

= 0.75 M x 0.5 L

= 3.75 mol

Molar mass of NaCl = 23 x 1 + 35.5 x 1

= 58.5 g / mol

Mass of NaCl = Molar mass x number of moles

= 58.5 g / mol x 3.75 mol

= 219.375 g

the correct answer would be b) volume versus temperature, because they are also directly proportional.