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How many mL of a 3.7 M HCI solution would it take to make a 250 mL of a 2.0 M solution? mL

How many grams of sucrose (molar mass is'342.30 g/mol) are in 528.1 mL of a 1.90 M solution? g sucrose A sodium hydroxide solution that contains 57.9 grams of NaOH per L of solution has a density of
1.14 g/mL. Calculate the molality of theNaOH in this solution. m NaOH
Calculate the concentration in ppm of lead in a solution that contains 179.8 mg Pb per 3.46 kg of
solution. ppm
What is the concentration of bromide, in ppm, if 152.4 g MgBr2 is dissolved in 3.51 L water. ppm of Br

User GrayedFox
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1 Answer

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Answer and Explanation:

To make a 250 mL of a 2.0 M solution from a 3.7 M HCl solution, you would need to use the formula M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Substituting the known values into the formula, we get:

3.7 M * V1 = 2.0 M * 250 mL

Solving for V1, we find that it would take approximately 135 mL of the 3.7 M HCl solution to make 250 mL of a 2.0 M solution.

To find out how many grams of sucrose are in 528.1 mL of a 1.90 M solution, you can use the formula moles = molarity * volume (in L). Since the molar mass of sucrose is 342.30 g/mol, you can then convert moles to grams using the formula mass = moles * molar mass.

First, convert the volume from mL to L: 528.1 mL * (1 L / 1000 mL) = 0.5281 L

Next, calculate the number of moles of sucrose in the solution: moles = 1.90 M * 0.5281 L = 1.00339 moles

Finally, convert moles to grams: mass = 1.00339 moles * 342.30 g/mol = 343.5 g

So, there are approximately 343.5 g of sucrose in 528.1 mL of a 1.90 M solution.

To calculate the molality of NaOH in a solution that contains 57.9 grams of NaOH per L of solution and has a density of 1.14 g/mL, you can use the formula molality = moles solute / kg solvent.

First, calculate the number of moles of NaOH in one liter of solution: moles = mass / molar mass = 57.9 g / 39.997 g/mol = 1.448 moles

Next, calculate the mass of one liter of solution: mass = density * volume = 1.14 g/mL * 1000 mL = 1140 g

Since water is the solvent and its density is approximately 1 g/mL, one liter of water has a mass of approximately 1000 g or 1 kg.

So, the molality of NaOH in this solution is: molality = moles solute / kg solvent = 1.448 moles / 1 kg = 1.448 mol/kg

To calculate the concentration in ppm (parts per million) of lead in a solution that contains 179.8 mg Pb per 3.46 kg of solution, you can use the formula ppm = (mass solute / mass solution) *10^6.

First, convert the mass of lead from mg to g: 179.8 mg * (1 g /1000 mg) = 0.1798 g

Next, substitute the known values into the formula: ppm = (mass solute / mass solution) *10^6 = (0.1798 g /3.46 kg) *10^6

So, the concentration of lead in this solution is approximately 51,965 ppm.

To calculate the concentration in ppm (parts per million) of bromide if 152.4 g MgBr2 is dissolved in 3.51 L water, you can use the formula ppm = (mass solute / mass solution) *10^6.

First, calculate the mass of bromide in MgBr2:

The molar mass of MgBr2 is 24.305 + (79.904 *2) =184.113 g/mol The mass fraction of Br in MgBr2 is (79.904*2)/184.113=0.867 The mass of Br in MgBr2 is (152/4)*0/867=132g

Next, calculate the mass of water:

The density of water is approximately 1g/mL, so its mass is approximately equal to its volume in milliliters. The mass of water is therefore approximately equal to (3/51L)*(1000mL/L)=3510g=3/51kg

Substitute these values into the formula for ppm:

ppm=(mass solute/mass solution)*10^6=(132g/3/51kg)*10^6=37/714ppm

So, there are approximately 37/714 ppm of bromide in the solution.

User JustcallmeDrago
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