What mass of glucose, #C_6H_12O_6#, would be required to prepare #5 * 10^3# #L# of a #0.215# #M# solution?

Answer 1

193661.25g #C_6H_12O_6# (before sig figs)

Molarity is defined as #M#= #(mols)/(1L solution)#
from this, we can deduct that .215#M# #C_6H_12O_6#= #(.215mol)/(1L solution)#
#C# = 12.01 * 6 = 72.06g #H# = 1.0079 * 12 = 12.0948g #O# = 16.00 * 6 = 96.00g #C_6H_12O_6#= 180.15g

Commence with what is provided.

#5*10^3##L# #C_6H_12O_6# #(.215 molC_6H_12O_6)/(1L solution)# #(180.15g)/(1mol C_6H_12O_6)#= 193661.25g #C_6H_12O_6# (before sig figs)
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Answer 2

To find the mass of glucose required, you can use the formula:

[ \text{Mass} = \text{Molarity} \times \text{Volume} \times \text{Molar Mass} ]

Where:

  • Molarity ((M)) = 0.215 M
  • Volume ((V)) = 5 * 10^3 L
  • Molar Mass of glucose ((M_\text{glucose})) = 180.16 g/mol

Substitute the given values:

[ \text{Mass} = 0.215 , \text{M} \times 5 * 10^3 , \text{L} \times 180.16 , \text{g/mol} ]

[ \text{Mass} ≈ 1939.2 , \text{g} ]

So, approximately 1939.2 grams of glucose would be required.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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