0.67 M
1.7M
32 g KCl
9.13
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
Dissolve 50 g of potassium carbonate in 100 mL of water at 20 0C.
This is a solution of 10 g KCl/100 g water.
an equilibrium between dissolved KCl and solid KCl
1.7M
32 g KCl
Molarity= Number of moles of solute/Liters of solution 50 grams KOH 700 ML to .7 Liters of h2o Molar Mass of KOH= 56 50 divided by 56 = .89 moles Molarity= .89 mol/.7 L = 1.27 MOLARITY
Molarity = moles of solute/Liters of solution Molarity = 25 moles sucrose/50 liters H2O = 0.5 M sucrose
m1v1 = m2v2 50 mL * 50 M = 25 mL * X M X = 100
Some conversion required. (mmolar into mol, or moles into mmol ) Molarity = moles of solute/Liters of solution 100 millimolar = 0.1 M glycine Molarity = moles of solute/Liters of solution manipulate algebraically Liters of solution = moles of solute/Molarity 0.005 mole glycine/0.1 M glycine = 0.05 Liters ( 1000 ml/1 L) = 50 milliliters of solution --------------------------------
6.0 M when you multiply 12M by the .50 liters you will get the 6.0 M
100 g water dissolve 45,8 g potassium chloride at 50 o 0C.
9.13
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