Asolution is made by dissolving 4.35 g of glucose (c6h1206) in 25.0 ml of water at 25 °c. calculate the molality of glucose in the solution. water has a density of 1.00 g/ml. suppose you take a solution and add more solvent, so that the original mass of solvent is doubled. you take this newsolution and add more solute, so that the original mass of the solute is doubled. what happens to the molality of the finalsolution, compared to the original molality? (a) it is doubled.(b) it is decreased by half.(c) it is unchanged.(d) it will increase or decrease depending on the molar mass of the solute.(e) there is no way to tell without knowing the molar mass of the solute.

The molality is unchanged (0.96 molal)

Explanation:

Step 1: Data given

mass of glucose = 4.35 grams

volume of water = 25.0 mL

Density of water = 1.00 g/mL

Molar mass of glucose = 180.156 g/mol

Step 2: Calculate number of moles

moles of glucose = mass of glucose / Molar mass of glucose

moles of glucose = 4.35 grams / 180.156 g/mol

moles of glucose = 0.024 moles

Step 3: Calculate mass of water

mass = density * volume

mass of water = 1.00 g/mL * 25.0 mL

mass of water = 25 g = 0.025 kg

Step 4: Calculate molality

molality = Number of moles / mass of water

molality = 0.024 moles / 0.025 kg

molality = 0.96 molal

Suppose you take a solution and add more solvent, so that the original mass of solvent is doubled.

This means double mass of water = 2*0.025 kg = 0.050 kg

Now molality is 0.024 moles / 0.050 kg = 0.48 molal

When the mass of solvent is doubled, the molality is halved from 0.96 molal to 0.48 molal

You take this newsolution and add more solute, so that the original mass of the solute is doubled.

This means double mass of glucose = 2*4.35 g = 8.70 g

8.70 grams of glucose = 8.7 grams * 180.156 g/mol = 0.048 moles

molality = 0.048 moles / 0.050 kg = 0.96 molal

The molality is unchanged

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.

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the political sponsor of weights and measures reform in the french revolutionary national assembly was the bishop of autun, better known as talleyrand. under his auspices, the french academy appointed several committees to carry out the work of developing a usable system of weights and measures for france. one of the committees recommended a decimalized measurement system based upon a length equal to one ten-millionth of the length of a quadrant of the earth’s meridian (i.e., one ten-millionth of the distance between the equator and the north pole).