1) How manygrams of solute are present in 57.0 ml of 250 Mk2Cr2O7? 2) If 4.26 g

Question

1)     How manygrams of solute are present in 57.0 ml of 250 Mk2Cr2O7?

2)    If 4.26 g of (NH4)2SO4 is dissolved in enough water toform 240 ml of solution, what is the molarity of thesolution?

3)      Howmany milliliters of 0.220 M CUSO4 contains 2.00 gof solute?

I am really havingdifficulty with molarity. Can someone help me with theseexamples so I can understand how to manipulate the equationsalgebraically? Thanks!

Explanation / Answer

1)     How many grams of solute are presentin 57.0 ml of 250 M k2Cr2O7? First, Lets figure out the molar mass of K2Cr2O7: Mm = (2*39) +(2*52) + (16*7) = 294 g K2Cr2O7 / 1 mole. We are given that we have a 250 Molar solution. This means we have250 moles K2Cr2O7 ( the solute) / 1L solution. Lets startthere, [250 mol K2Cr2O7 ] * [ 294 gK2Cr2O7] = 73500 g K2Cr2O7 in 1L of oursolution. Now lets convert to mL. [    1L solution        ]    [ 1 mol K2Cr2O7] [73500 g K2Cr2O7 ] * [    1L     ] = 73.5 g K2Cr2O7 (solute) per 1 mL ofsolution. [ 1 Lsolution           ]    [1000mL] in 57 mL of solution you would have: (57 mL solution) ( 73.5 gsolute / 1 mL solution) = 4200g! You could certainly combine the steps to save time, but whenworking these problems its always a good idea to start with theMolar mass and converting to what you want by getting your units tocancel. 2. If 4.26 g of (NH4)2SO4 is dissolved in enough water toform 240 ml of solution, what is the molarity of thesolution? Molar mass of (NH4)2SO4 = 132 g/mol how many moles do we have? --> 4.26g(NH4)2SO4 divided by the molar mass gives us molesof (NH4)2SO4. I get .032 moles of (NH4)2SO4. Molarity is moles of soluteper liter of solution. Great! all thats left is to convert 240 mLto liters. [240mL] * [    1 L ] =.24L                    So our molarity is .032 mol (NH4)2SO4 per .24 L. or ~.13M                  [1000 mL] 3)      How many millilitersof 0.220 M CUSO4 contains 2.00 g of solute? Molar mass of CuSO4 = 159.5g/mol. Our solution is .220M. Write this down first and then lets convertto grams: [.220 mol CUSO4 ] * [159.5 gCUSO4] = 35.09 g CUSO4 [ 1 Lsolution        ]   [ 1 mol CUSO4  ]        1 Lsolution But we want milliliters, lets convert: 35.09 gCUSO4 * [    1L       ] = .03509gCuSO4 per 1 mL. Thisis the amount of g CuSO4 in 1 mL, so      1 L    [ 1000 mL] 2.00g CuSO4 = .03509 gCuSO4      xmL                     1 mL solve the porportion for x for the amount of mL solution containing2.00g CuSO4. ~ 57.0 mL

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