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fomenos
3 years ago
15

A solution is prepared by dissolving 50.4 gg sucrose (C12H22O11C12H22O11) in 0.332 kgkg of water. The final volume of the soluti

on is 355 mLmL. For this solution, calculate the molarity.
Chemistry
1 answer:
elena-s [515]3 years ago
4 0

Answer:

The correct answer is 0.41 M

Explanation:

The molarity of a solution is given by the number of moles of solute per liter of solution.

Molarity= moles solute/1 L solution= mol/L

First we need the molecular weight (Mw) of sucrose in order to convert the mass of sucrose (in g) to mol. We can quickly calculate this from the molar mass (MM) of the elements C, H and O in Periodic Table:

Mw sucrose (C₁₂H₂₂O₁₁)= (12 x MM C) +(22 x MM H) + (11 x MM O)= (12 x 12 g/mol)+(22 x 1 g/mol) + (11 x 16 g/mol)= 342 g/mol

Then, we divide the mass of sucrose (50.4 g) into the molecular weight of sucrose to obtain the number of moles and this is divided into the volume of the solution (355 ml). We convert ml to L taking into account that 1000 ml= 1L:

50.4 g sucrose x 1 mol/342 g sucrose x 1/355 ml x 1000 ml/1 L= 0.415 mol/L= 0.41 M

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5 0
3 years ago
How many milliliters of 0.100 m naoh are required to neutralize 9.00 ml of 0.0500 m hcl?
BabaBlast [244]
V  ( NaOH ) = mL ?

M ( NaOH ) = 0.100 M

V ( HCl ) = 9.00 mL / 1000 => 0.009 L

M ( HCl ) = 0.0500 M

number of moles HCl:

n = M x V

n = 0.009 x 0.0500 => 0.00045 moles HCl

mole ratio:

<span>HCl + NaOH = NaCl + H2O
</span>
 1 mole HCl ---------------- 1 mole NaOH
 0.00045 moles HCl ----- ??

0.00045 x 1 / 1 => 0.00045 moles of NaOH

M = n / V

0.100 = 0.00045 / V

V = 0.00045 / 0.100

V = 0.0045 L

1 L ------------ 1000 mL
0.0045 L ----- ??

0.0045 x 1000 / 1 => 4.5 mL of NaOH



6 0
3 years ago
What is the change in the freezing point of water when 68.5 g of sucrose is
mr Goodwill [35]
The change is that the water will freeze to 0 or minus I don’t know as I’m not to sure
7 0
3 years ago
Read 2 more answers
Cho recorded the weather in her town for a week. She wrote down her observations in the table below. Mon Tue Wed Thu Fri Sat Sun
SIZIF [17.4K]

Answer:

If this trend continues, the following week will be cooler, and a large amount of rain will fall.

Explanation:

Patterns and trends can often be found in data sets. During the week that Cho recorded the weather, the temperatures consistently dropped by one to four degrees each day. At the end of the week, the amount of precipitation increased daily.

6 0
3 years ago
How many moles are in 3.45 g of Ba(CIO3)2 ?
topjm [15]
<h3>Answer:</h3>

0.0113 mol Ba(ClO₃)₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structures</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.45 g Ba(ClO₃)₂

<u>Step 2: Identify Conversions</u>

Molar Mass of Ba - 137.33 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Ba(ClO₃)₂ - 137.33 + 2(35.45) + 6(16.00) = 304.33 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.45 \ g \ Ba(ClO_3)_2(\frac{1 \ mol \ Ba(ClO_3)_2}{304.33 \ g \ Ba(ClO_3)_2})
  2. Multiply/Divide:                \displaystyle 0.011336 \ mol \ Ba(ClO_3)_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.011336 mol Ba(ClO₃)₂ ≈ 0.0113 mol Ba(ClO₃)₂

8 0
3 years ago
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