Given: 0.5 g NaCl; 0.05 L of solution.1) Find the molar mass of NaCl.(22.99) + (35.45) = 58.44 g/mol of NaCl2) Convert grams of NaCl to moles of NaCl.(0.5 g NaCl) X (1 mol NaCl / 58.44 g NaCl) = 0.00855578 mol NaCl3) Use the molarity equation, M = mol / L, to solve for molarity (M).(0.00855578 mol NaCl / 0.05 L) = 0.17 M

What is the molarity of a solution containing 325 g of NaCl dissolved in 750.ml of solution?

0,1 M NaCl is equal to 0,5844 g/L.

The answer is 0,514 mol.

2 percent NaCl is 2 g NaCl/100g water or 20 g NaCl/1000 water.

Find moles NaCl first.14.60 grams NaCl (1 mole NaCl/58.44 grams)= 0.2498 moles NaCl================Now,Molarity = moles of solute/Liters of solutionMolarity = 0.2498 moles NaCl/2.000 Liters= 0.1249 M NaCl solution--------------------------------

Molarity = moles of solute/volume of solution so, 125 grams of NaCl (1mol NaCl/58.44 grams NaCl) = 2.1389 moles NaCl/400 Liters = 0.00535 M NaCl

Molarity = moles of solute/Liters of solution Find moles NaCl 9 grams NaCl (1 mole NaCl/58.44 grams) = 0.154 moles NaCl Molarity = 0.154 moles NaCl/1 Liter = 0.2 M sodium chloride -------------------------------

10 g of NaCl in 2 kg of water is 5 g of NaCl in 1 kg of water.The molar mass of NaCl is 39,666 g.5 g is equal to 0,318 molal.

5 % NaCl is equal to 5 g NaCl in 100 g of a material.

Pretty concentrated! ( I assume you mean 2.3 Liters )First get the moles of sodium chloride.456 grams NaCl (1 mole NaCl/58.44 grams)= 7.803 moles NaCl==============Now,Molarity = moles of solute/Liters of solutionMolarity = 7.803 moles NaCl/2.3 Liters= 3.4 M NaCl solution----------------------------

Concentration = Molarity = mol/L24 g NaCl = ?? mol NaCl?? mol NaCl/2 L water = ?? M (M is unit of molarity)

The density is 1,0075 g/cm3 at 20 0C.

Almost exactly 1 M - to be exactly 1.0M would require 58.5 g NaCl

Dissolve 116,879 g of pure NaCl in 1 L demineralized water at 20 0C.

Divide grams (mass) by molar mass to find moles58.44 (g NaCl/L) / [22.99+35.45](g NaCl/mol NaCl)= 1.000 mol/L NaCl

10 percent NaCl in distilled water is equivalent to 100 g/L NaCl.

125g NaCl x 1 mol NaCl / 58.44 NaCl= 2.13894 mol / 4.00 L = 0.535 M

Molarity = moles solute/liter solutionmoles solute = 7 g NaCl x 1 mol NaCl/58 g NaCl = 0.12 moles NaClliters of solution = 450.0 ml x 1 L/1000 ml = 0.450 litersmolarity = 0.12 moles/0.450 liters = 0.268 M = 0.3 M (one sig fig)

The answer is 8 g NaCl.

If you think to an isotonic solution the concentration is 0,9 g NaCl/100 g solution.

Get moles NaCl and change 245 ml to 0.245 Liters. 3.8 grams NaCl (1 mole NaCl/58.54 grams) = 0.0650 moles NaCl Molarity = moles of solute/Liters of solution Molarity = 0.0650 moles NaCl/0.245 Liters = 0.27 M NaCl ----------------------

The answer is 20 g NaCl in 1 L solution.

This concentration of NaCl is 2,6 g NaCl/100 mL solution.

Molarity of a solution is defined as the number of moles, or for ionic compounds such as NaCl, gram formula masses, of solute contained in exactly 1 liter of solution. The gram formula mass of NaCl is 58.44, and 725 mL is 0.725 liter. The molarity is therefore [(23.1/58.44)/0.725] or 0.545, to the justified number of significant digits.

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