Best Answer

Multiply the molarity (M, which is in mol/L) with the volume (in L) to get the number of moles needed. Then multiply the result with the molar mass. If you look at the units they will cancel to give an answer in grams.

(mol/L)*(L)=mol, (mol)*(g/mol)=g

So for the numerical answer you get (0.0552 mol/L)*(0.750 L)*(119.00 g/mol)= 4.93 g KBr

Q: How many grams of KBr are required to make 750.0 mL of solution that is 0.0552M molar mass of KBr 119.00gmol?

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30 grams

40 grams, this is the 1M NaOH standard laboratory solution.

400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.

200 grams of solution will contain 200 x 4% or 200 x 0.04 = 8.0 total grams of solute.

This is a homogeneous solution of ammonium hydroxide in water.

Related questions

30 grams

3.33

122.5g

40 grams, this is the 1M NaOH standard laboratory solution.

You need 0,9 glucose.

Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed

400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.

mixture

489 grams

The percent concentration is 13,75 %.

the formula is no. moles is mass / molecular mass. As the number of moles is 1, the mass required will be exactly the same as the molecular mass, which is 58.32g

The molecular weight of NaCl is 58.44; sodium =22.99; Chlorine=35.45. A 1 molar solution is the molecular weight in grams in 1 litre of water, so a 3.5 molar solution would be 58.44g multiplied by 3.5, which is 204.54g in 1L.